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Stoichiometry Made Simple

This interactive provides step-by-step explanations and examples of many different types of stoichiometry problems. The diagram below shows the possible "paths" that can be taken to get from one quantity to another.

Mass of A to Moles of B

Mass of A to Moles of A

Select a green starting position on the diagram.

Volume of A to Volume of B

volume of gas A (liters)

Moles of A to Mass of A

Particles of A to Particles of B

volume of gas B (liters)

Moles of A to Particles of A

Use balanced equation

Mass of A to Mass of B

Particles of A to Moles of A

Volume of Gas A to Moles of A

Moles of A to Volume of Gas A

1 Mole = 22.4 L of gas at STP

Mass of A to Particles of A

Particles of A to Mass of A

mass A (grams)

moles A

Volume of Gas A to Particles of A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

Particles of A to Volume of Gas A

Mass of Gas A to Volume of Gas A

1 Mole = 6.02 x 1023 particles

Volume of Gas A to Mass of Gas A

particles A

Use balanced equation

particles B

Brought to you by Virtual Science Teachers

Stoichiometry /ˌstɔɪkiˈɒmɪtri/ refers to the relationship between the quantities of reactants and products before, during, and following chemical reactions.

Stoichiometry Made Simple

What kind of problem would you like to solve?

Mass of A to Moles of B

Mass of A to Moles of A

Volume of A to Volume of B

Moles of A to Mass of A

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

Particles of A to Particles of B

Moles of A to Particles of A

Mass of A to Mass of B

Particles of A to Moles of A

Volume of Gas A to Moles of A

Moles of A to Volume of Gas A

Mass of A to Particles of A

Particles of A to Mass of A

Volume of Gas A to Particles of A

Particles of A to Volume of Gas A

Mass of Gas A to Volume of Gas A

Volume of Gas A to Mass of Gas A

Brought to you by Virtual Science Teachers

Stoichiometry Made Simple

What kind of problem would you like to solve?

Mass of A to Moles of B

Mass of A to Moles of A

Volume of A to Volume of B

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

Moles of A to Mass of A

Particles of A to Particles of B

Moles of A to Particles of A

Mass of A to Mass of B

Particles of A to Moles of A

Mass of A to Volume of B

Volume of Gas A to Moles of A

Mass of A to Particles of B

Moles of A to Volume of Gas A

Mass of A to Moles of B

Mass of A to Particles of A

Volume of A to Moles of B

Particles of A to Mass of A

Volume of A to Mass of B

Volume of Gas A to Particles of A

Particles of A to Volume of Gas A

Mass of Gas A to Volume of Gas A

Volume of Gas A to Mass of Gas A

Brought to you by Virtual Science Teachers

Stoichiometry Made Simple

Any path that leads to a blue circle involves the quantities of products and reactants involved in a chemical reaction.

Mass of A to Moles of B

Mass of A to Moles of A

Now select an ending position on the diagram.

Volume of A to Volume of B

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

Moles of A to Mass of A

Particles of A to Particles of B

Moles of A to Particles of A

Mass of A to Mass of B

Particles of A to Moles of A

Volume of Gas A to Moles of A

Moles of A to Volume of Gas A

Mass of A to Particles of A

Particles of A to Mass of A

Volume of Gas A to Particles of A

Particles of A to Volume of Gas A

Mass of Gas A to Volume of Gas A

Volume of Gas A to Mass of Gas A

Brought to you by Virtual Science Teachers

Stoichiometry Made Simple

Any path that leads to a blue circle involves the quantities of products and reactants involved in a chemical reaction.

Now select an ending position on the diagram.

Mass of A to Moles of B

Mass of A to Moles of A

Volume of A to Volume of B

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

Moles of A to Mass of A

Particles of A to Particles of B

Moles of A to Particles of A

Mass of A to Mass of B

Particles of A to Moles of A

Volume of Gas A to Moles of A

Moles of A to Volume of Gas A

Mass of A to Particles of A

Particles of A to Mass of A

Volume of Gas A to Particles of A

Particles of A to Volume of Gas A

Mass of Gas A to Volume of Gas A

Volume of Gas A to Mass of Gas A

Brought to you by Virtual Science Teachers

Coming Soon

Stoichiometry Made Simple

Any path that leads to a blue circle involves the quantities of products and reactants involved in a chemical reaction.

Mass of A to Moles of B

Now select an ending position on the diagram.

Mass of A to Moles of A

Volume of A to Volume of B

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

Moles of A to Mass of A

Particles of A to Particles of B

Moles of A to Particles of A

Mass of A to Mass of B

Particles of A to Moles of A

Volume of Gas A to Moles of A

Moles of A to Volume of Gas A

Mass of A to Particles of A

Particles of A to Mass of A

Coming Soon

Volume of Gas A to Particles of A

Particles of A to Volume of Gas A

Mass of Gas A to Volume of Gas A

Volume of Gas A to Mass of Gas A

Brought to you by Virtual Science Teachers

Coming Soon

Stoichiometry Made Simple

Coming Soon

Any path that leads to a blue circle involves the quantities of products and reactants involved in a chemical reaction.

Mass of A to Moles of B

Now select an ending position on the diagram.

Mass of A to Moles of A

Volume of A to Volume of B

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

Moles of A to Mass of A

Particles of A to Particles of B

Moles of A to Particles of A

Mass of A to Mass of B

Particles of A to Moles of A

Volume of Gas A to Moles of A

Moles of A to Volume of Gas A

Mass of A to Particles of A

Coming Soon

Particles of A to Mass of A

Volume of Gas A to Particles of A

Particles of A to Volume of Gas A

Mass of Gas A to Volume of Gas A

Volume of Gas A to Mass of Gas A

Brought to you by Virtual Science Teachers

106 grams NaCl = ? moles NaCl

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

58.4 grams NaCl

1 mole NaCl

106 grams NaCl

1.82 moles NaCl

Step 1: Put what you start with over 1.

106 grams NaCl

Step 2: Draw a multiplication sign and another line.

106 grams NaCl

Step 4: Write the units of your next stop in the numerator.

grams NaCl

mole NaCl

106 grams NaCl

Step 5: Use the corresponding conversion factor to write in the values.

58.4 grams NaCl

1 mole NaCl

106 grams NaCl

1 mole of NaCl has a mass of 58.4 grams.

106 grams NaCl

Step 3: Write the units you want to cancel out the denominator of this new fraction.

grams NaCl

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

*Note: This simulation shows the immediate water displacement when a rainbow swirl is added to water, before it dissolves.

mass (grams)

Step 6: Cancel out units that are both in the numerator and denominator and multiply.

25.3 moles H20 = ? grams H20

1 mole H2O

18.0 grams H2O

25.3 moles H20

455 grams H20

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

Step 1: Put what you start with over 1.

25.3 moles H20

Step 4: Write the units of your next stop in the numerator.

mole H2O

grams H20

25.3 moles H2O

1 mole of H2O has a mass of 18.0 grams.

Step 5: Use the corresponding conversion factor to write in the values.

1 mole H2O

18.0 grams H2O

25.3 moles H20

25.3 moles H2O

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

mole H20

Step 2: Draw a multiplication sign and another line.

25.3 moles H20

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

*Note: This simulation shows the immediate water displacement when a rainbow swirl is added to water, before it dissolves.

Step 6: Cancel out units that are both in the numerator and denominator and multiply.

20.0 moles H20 = ? molecules H20

20.0 moles H20

1 mole H2O

6.02 x 1023 molecules H2O

1.20 x 1024 molecules H20

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

Step 5: Use the corresponding conversion factor to write in the values.

1 mole H2O

6.02 x 1023 molecules H2O

20.0 moles H20

1 mole of H2O has 6.02 x 1023 molecules.

Step 4: Write the units of your next stop in the numerator.

mole H2O

molecules H20

20.0 moles H2O

Step 2: Draw a multiplication sign and another line.

20.0 moles H20

20.0 moles H2O

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

mole H20

Step 1: Put what you start with over 1.

20.0 moles H20

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

*Note: This simulation shows the immediate water displacement when a rainbow swirl is added to water, before it dissolves.

Step 6: Cancel out units that are both in the numerator and denominator and multiply.

1.29 x 1024 Na atoms = ? moles of Na

2.14 moles Na

6.02 x 1023 Na atoms

1 mole Na

1.29 x 1024 Na atoms

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

Step 5: Use the corresponding conversion factor to write in the values.

6.02 x 1023 Na atoms

1 mole of Na has 6.02 x 1023 atoms.

1 mole Na

1.29 x 1024 Na atoms

Step 2: Draw a multiplication sign and another line.

1.29 x 1024 Na atoms

Step 1: Put what you start with over 1.

1.29 x 1024 Na atoms

Step 4: Write the units of your next stop in the numerator.

mole Na

1.29 x 1024 Na atoms

Na atoms

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

1.29 x 1024 Na atoms

Na atoms

20 moles H20

mole H20

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

20 moles H2O

mole H2O

20 moles H20

1.20 x 1024 molecules H20

*Note: This simulation shows the immediate water displacement when a rainbow swirl is added to water, before it dissolves.

5.60 Liters CH4 at STP= ? moles of CH4

0.250 moles CH4

1 mole CH4

22.4 L CH4

5.60 L CH4

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

Step 1: Put what you start with over 1.

5.60 L CH4

Step 5: Use the corresponding conversion factor to write in the values.

1 mole of any gas at STP has a volume of 22.4 L.

1 mole CH4

22.4 L CH4

5.60 L CH4

Step 4: Write the units of your next stop in the numerator.

mole CH4

L CH4

5.60 L CH4

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

L CH4

5.60 L CH4

Step 2: Write a multiplication sign and another line.

5.60 L CH4

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

Step 6: Cancel out units that are both in the numerator and denominator and multiply.

12.0 moles of O2= ? Liters of O2 at STP

269 L O2

22.4 L O2

12.0 moles O2

1 mole O2

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

Step 1: Put what you start with over 1.

12.0 moles O2

Step 5: Use the corresponding conversion factor to write in the values.

1 mole of any gas at STP has a volume of 22.4 L.

22.4 L O2

12.0 moles O2

1 mole O2

Step 4: Write the units of your next stop in the numerator.

L O2

12.0 moles O2

mole O2

Step 3: Write the units you to cancel out in the denominator of this new fraction.

moles O2

12.0 moles O2

Step 2: Draw a multiplication sign and another line.

12.0 moles O2

1.29 x 1024 Na atoms

20 moles H20

mole H20

20 moles H2O

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

moles O2

1.29 x 1024 Na atoms

L CH4

20 moles H2O

Na atoms

22.4 L CH4

mole H2O

20 moles H20

6.02 x 1023 Na atoms

20 moles H20

1.20 x 1024 molecules H20

*Note: This simulation shows the immediate water displacement when a rainbow swirl is added to water, before it dissolves.

Step 6: Cancel out units that are both in the numerator and denominator and multiply.

30.0 grams of H2O = ? Molecules of H2O

1 mole H2O

30.0 g H2O

18.0 g H2O

1 mole H2O

6.02 x 1023 molecules H2O

1.00 x 1024 molecules H2O

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

1 mole H2O

30.0 g H2O

18.0 g H2O

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole H2O

6.02 x 1023 molecules H2O

1 mole of H2O has 6.02 x 1023 molecules.

Repeat Step 4: Write the units of your next stop in the numerator.

1 mole H2O

30.0 g H2O

18.0 g H2O

mole H2O

molecules H2O

Step 1: Put what you start with over 1.

30.0 g H2O

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

g H2O

Step 4: Write the units of your next stop in the numerator.

mole H2O

30.0 g H2O

g H2O

Step 2: Write a multiplication sign and another line.

30.0 g H2O

Step 5: Use the corresponding conversion factor to write in the values.

1 mole of H2O has a mass of 18.0 grams.

1 mole H2O

30.0 g H2O

18.0 g H2O

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

1 mole H2O

30.0 g H2O

18.0 g H2O

mole H2O

Repeat Step 2: Write a multiplication sign and another line.

1 mole H2O

30.0 g H2O

18.0 g H2O

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

Step 6: Cancel out units that are both in the numerator and denominator and multiply.

95.0 x 1024 K atoms = ? grams K

95.0 x 1023 K atoms = ? grams K

617 g K

1 mole K

95.0 x 1023 K atoms

6.02 x 1023 K atoms

1 mole K

39.1 g K

Step 6: Cancel out units that are both in the numerator and denominator. Then, multiply the top numbers and divide the bottom numbers to get your answer!

Step 1: Put what you start with over 1.

95.0 x 1023 K atoms

Repeat Step 4: Write the units of your next stop in the numerator.

g K

1 mole K

95.0 x 1023 K atoms

6.02 x 1023 K atoms

mole K

Step 2: Draw a multiplication sign and another line.

95.0 x 1023 K atoms

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole of K has a mass of 39.1 grams.

1 mole K

95.0 x 1023 K atoms

6.02 x 1023 K atoms

1 mole K

39.1 g K

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

1 mole K

95.0 x 1023 K atoms

6.02 x 1023 K atoms

mole K

Step 5: Use the corresponding conversion factor to write in the values.

Repeat Step 2: Draw a multiplication sign and another line.

1 mole K

95.0 x 1023 K atoms

1 mole of K contains 6.02 x 1023 K atoms

6.02 x 1023 K atoms

1 mole K

95.0 x 1023 K atoms

6.02 x 1023 K atoms

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

Step 4: Write the units of your next stop in the numerator.

mole K

95.0 x 1023 K atoms

K atoms

95.0 x 1023 K atoms

K atoms

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

1 Mole = 22.4 L of gas at standard temperature & pressure

mole H2O

Step 6: Cancel out units that are both in the numerator and denominator and multiply.

36.0 x 1023 CO2 molecules = ? Liters CO2 at STP

134 L CO2

1 mole CO2

22.4 L CO2

1 mole CO2

34.0 x 1023 CO2 molecules

6.02 x 1023 CO2 molecules

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

mole CO2

1 mole CO2

36.0 x 1023 CO2 molecules

6.02 x 1023 CO2 molecules

mole CO2

Repeat Step 4: Write the units of your next stop in the numerator.

L CO2

1 mole CO2

36.0 x 1023 CO2 molecules

6.02 x 1023 CO2 molecules

1 mole of any gas at STP has a volume of 22.4 L.

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole CO2

22.4 L CO2

1 mole CO2

34.0 x 1023 CO2 molecules

6.02 x 1023 CO2 molecules

Step 4: Write the units of your next stop in the numerator.

mole CO2

36.0 x 1023 CO2 molecules

CO2 molecules

Step 5: Use the corresponding conversion factor to write in the values.

1 mole of CO2 contains 6.02 x 1023 molecules.

1 mole CO2

36.0 x 1023 CO2 molecules

6.02 x 1023 CO2 molecules

Step 1: Put what you start with over 1.

36.0 x 1023 CO2 molecules

Repeat Step 2: Draw a multiplication sign and another line.

1 mole CO2

36.0 x 1023 CO2 molecules

6.02 x 1023 CO2 molecules

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

36.0 x 1023 CO2 molecules

CO2 molecules

Step 2: Draw a multiplication sign and another line.

36.0 x 1023 CO2 molecules

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

Step 6: Cancel out units that are both in the numerator and denominator and multiply.

7.50 Liters of CH4 at STP = ? Molecules CH4

2.02 x 1023 molecules CH4

6.02 x 1023 molecules CH4

1 mole CH4

22.4 L CH4

7.50 L CH4

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

Repeat Step 4: Write the units of your next stop in the numerator.

molecules CH4

mole CH4

1 mole CH4

22.4 L CH4

7.50 L CH4

Step 5: Use the corresponding conversion factor to write in the values.

1 mole of any gas at STP has a volume of 22.4 L.

1 mole CH4

22.4 L CH4

7.50 L CH4

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

mole CH4

1 mole CH4

22.4 L CH4

7.50 L CH4

Step 4: Write the units of your next stop in the numerator.

mole CH4

L CH4

7.50 L CH4

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

L CH4

7.50 L CH4

Repeat Step 2: Draw a multiplication sign and another line.

1 mole CH4

22.4 L CH4

7.50 L CH4

There are 6.02 x 1023 molecules in a mole of CH4.

Repeat Step 5: Use the corresponding conversion factor to write in the values.

6.02 x 1023 molecules CH4

1 mole CH4

22.4 L CH4

7.50 L CH4

Step 2: Draw a multiplication sign and another line.

7.50 L CH4

Step 1: Put what you start with over 1.

7.50 L CH4

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

22.4 L CO2

1 mole CO2

36.0 x 1023 CO2 molecules

6.02 x 1023 CO2 molecules

36.0 x 1023 CO2 molecules

mole CO2

CO2 molecules

Step 6: Cancel out units that are both in the numerator and denominator and multiply.

61.5 grams CO at STP = ? Liters CO

49.2 L CO

1 mole CO

28.0 g CO

61.5 g CO

22.4 L CO

1 mole CO

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

mole CO

1 mole CO

28.0 g CO

61.5 g CO

Step 1: Put what you start with over 1.

61.5 g CO

Repeat Step 4: Write the units of your next stop in the numerator.

L CO

mole CO

1 mole CO

28.0 g CO

61.5 g CO

Step 5: Use the corresponding conversion factor to write in the values.

1 mole of CO has a mass of 28.0 grams.

1 mole CO

28.0 g CO

61.5 g CO

One mole of any gas at STP has a volume of 22.4 L.

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole CO

28.0 g CO

61.5 g CO

22.4 L CO

1 mole CO

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

g CO

61.5 g CO

Step 4: Write the units of your next stop in the numerator.

mole CO

g CO

61.5 g CO

Repeat Step 2: Draw a multiplication sign and another line.

1 mole CO

28.0 g CO

61.5 g CO

Step 2: Draw a multiplication sign and another line.

61.5 g CO

Use balanced equation

1 Mole = 6.02 x 1023 particles

particles B

1 Mole = 6.02 x 1023 particles

mass A (grams)

Use Molar Mass

particles A

volume of gas A (liters)

moles B

Use balanced equation

1 Mole = 22.4 L of gas at STP

moles A

mass B (grams)

Use Molar Mass

volume of gas B (liters)

1 Mole = 22.4 L of gas at STP

Use balanced equation

28.0 g CO

61.5 g CO

28.0 g CO

61.5 g CO

28.0 g CO

61.5 g CO

Step 6: Cancel out units that are both in the numerator and denominator and multiply.

61.5 g CO

44.8 Liters N2 at STP = ? grams N2

1 mole N2

56.0 g N2

44.8 L N2

22.4 L N2

28.0 g N2

1 mole N2

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

Repeat Step 4: Write the units of your next stop in the numerator.

g N2

mole N2

1 mole N2

44.8 L N2

22.4 L N2

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

mole N2

1 mole N2

28.0 g CO

44.8 L N2

Step 1: Put what you start with over 1.

44.8 L N2

Step 2: Draw a multiplication sign and another line.

44.8 L N2

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

44.8 L N2

L N2

Repeat Step 2: Draw a multiplication sign and another line.

1 mole N2

44.8 L N2

22.4 L N2

Step 5: Use the corresponding conversion factor to write in the values.

1 mole N2

One mole of any gas at STP has a volume of 22.4 L.

44.8 L N2

22.4 L N2

28.0 g N2

1 mole of N2 has a mass of 28.0 grams.

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole N2

44.8 L N2

22.4 L N2

Step 4: Write the units of your next stop in the numerator.

mole CO

44.8 L N2

L N2

61.5 g CO

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

61.5 g CO

1 Mole = 22.4 L of gas at STP

61.5 g CO

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

1 Mole = 6.02 x 1023 particles

Use balanced equation

particles A

particles B

61.5 g CO

g CO

28.0 g CO

g CO

61.5 g CO

22.4 L CO

1 mole CO

28.0 g CO

L N2

2KClO3 → 2KCl + 3O2

Repeat Step 2: Draw a multiplication sign and another line.

1 mole N2

44.8 L N2

How many moles of O2 are produced when 10.6 moles of KClO3 decompose?

22.4 L N2

61.5 g CO

= 15.9 moles O2

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

10.6 moles KClO3

2 moles KClO3

3 moles O2

Step 2: Draw a multiplication sign and another line.

10.6 moles KClO3

Step 1: Put what you start with over 1.

10.6 moles KClO3

Step 5: Use the corresponding conversion factor to write in the values.

The balanced reaction shows the ratio to O2 to KClO3 is 3:2.

10.6 moles KClO3

2 moles KClO3

3 moles O2

2KClO3 → 2KCl + 3O2

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

10.6 moles KClO3

moles KClO3

Step 4: Write the units of your next stop in the numerator.

10.6 moles KClO3

moles KClO3

moles O2

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

44.8 L N2

1 mole N2

61.5 g CO

28.0 g CO

mole N2

1 Mole = 22.4 L of gas at STP

Repeat Step 4: Write the units of your next stop in the numerator.

1 mole N2

61.5 g CO

g N2

44.8 L N2

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

22.4 L N2

mole N2

1 Mole = 6.02 x 1023 particles

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole N2

28.0 g N2

44.8 L N2

Use balanced equation

particles A

particles B

1 mole N2

22.4 L N2

1 mole of N2 has a mass of 28.0 grams.

61.5 g CO

g CO

28.0 g CO

g CO

61.5 g CO

22.4 L CO

1 mole CO

Repeat Step 2: Draw a multiplication sign and another line.

1 mole N2

44.8 L N2

22.4 L N2

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

44.8 L N2

1 mole N2

28.0 g CO

mole N2

Repeat Step 4: Write the units of your next stop in the numerator.

1 mole N2

g N2

44.8 L N2

22.4 L N2

mole N2

28.0 g CO

10.6 moles KClO3

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole N2

28.0 g N2

44.8 L N2

10.6 moles KClO3

1 mole N2

22.4 L N2

1 mole of N2 has a mass of 28.0 grams.

L N2

C3H8 + 5O2 → 3CO2 + 4H2O

How many liters of O2 are needed for the complete combustion of 11.5 liters of C3H8?

61.5 g CO

= 57.5 L O2

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

1 L C3H8

11.5 L C3H8

5 L O2

Step 1: Put what you start with over 1.

11.5 L C3H8

Step 2: Draw a multiplication sign and another line.

11.5 L C3H8

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

L C3H8

11.5 L C3H8

Step 5: Use the corresponding conversion factor to write in the values.

The balanced reaction shows the ratio to O2 to C3H8 is 5:1.

1C3H8 + 5O2 → 3CO2 + 4H2O

1 L C3H8

11.5 L C3H8

5 L O2

Step 4: Write the units of your next stop in the numerator.

L C3H8

11.5 L C3H8

L O2

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

61.5 g CO

2KClO3 → 2KCl + 3O2

1 Mole = 22.4 L of gas at STP

61.5 g CO

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

1 Mole = 6.02 x 1023 particles

Use balanced equation

particles A

particles B

61.5 g CO

g CO

2000 SOUNDINGS CRESCENT COURT, SUFFOLK, VA 23435

28.0 g CO

g CO

61.5 g CO

22.4 L CO

1 mole CO

Repeat Step 2: Draw a multiplication sign and another line.

1 mole N2

44.8 L N2

22.4 L N2

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

44.8 L N2

1 mole N2

28.0 g CO

mole N2

Repeat Step 4: Write the units of your next stop in the numerator.

1 mole N2

g N2

44.8 L N2

22.4 L N2

mole N2

28.0 g CO

10.6 moles KClO3

Repeat Step 5: Use the corresponding conversion factor to write in the values.

11.5 L C3H8

1 mole N2

28.0 g N2

44.8 L N2

10.6 moles KClO3

1 mole N2

22.4 L N2

1 mole of N2 has a mass of 28.0 grams.

L N2

1C3H8 + 5O2 → 3CO2 + 4H2O

3H2 + N2 → 2NH3

How many molecules of NH3 are produced when 45.8 x 1023 molecules of H2 completely react with N2?

61.5 g CO

= 30.5 x 1024 molecules NH3

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

2 molecules NH3

3 molecules H2

45.8 x 1024 molecules H2

Step 1: Put what you start with over 1.

45.8 x 1024 molecules H2

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

molecules H2

45.8 x 1024 molecules H2

Step 2: Draw a multiplication sign and another line.

45.8 x 1024 molecules H2

Step 4: Write the units of your next stop in the numerator.

molecules NH3

molecules H2

45.8 x 1024 molecules H2

Step 5: Use the corresponding conversion factor to write in the values.

The balanced reaction shows the ratio to NH3 to H2 is 2:3.

2 molecules NH3

3 molecules H2

45.8 x 1024 molecules H2

3H2 + N2 → 2NH3

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

61.5 g CO

2KClO3 → 2KCl + 3O2

1 Mole = 22.4 L of gas at STP

61.5 g CO

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

1 Mole = 6.02 x 1023 particles

Use balanced equation

particles A

particles B

61.5 g CO

g CO

2000 SOUNDINGS CRESCENT COURT, SUFFOLK, VA 23435

28.0 g CO

g CO

61.5 g CO

22.4 L CO

1 mole CO

28.0 g CO

61.5 g CO

mole N2

28.0 g CO

61.5 g CO

g N2

28.0 g CO

61.5 g CO

Step 6: Cancel out units that are both in the numerator and denominator and multiply.

61.5 g CO

mole KClO3

44.8 Liters N2 at STP = ? grams N2

2KClO3 → 2KCl + 3O2

How many grams of O2 are produced when 96.2 grams of KClO3 decompose?

= 37.7 g O2

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

1 mole O2

2 mole KClO3

3 mole O2

1 mole KClO3

122.6 g KClO3

96.2 g KClO3

32.0 g O2

61.5 g CO

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

1 mole KClO3

122.6 g KClO3

96.2 g KClO3

mole KClO3

Repeat Step 4: Write the units of your next stop in the numerator.

mole KClO3

1 mole KClO3

122.6 g KClO3

96.2 g KClO3

mole O2

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

1 mole KClO3

122.6 g KClO3

96.2 g KClO3

2 mole KClO3

3 mole O2

mole O2

Repeat Step 4: Write the units of your next stop in the numerator.

mole O2

2 mole KClO3

3 mole O2

1 mole KClO3

122.6 g KClO3

96.2 g KClO3

g O2

Step 4: Write the units of your next stop in the numerator.

mole KClO3

g KClO3

96.2 g KClO3

Repeat Step 2: Draw a multiplication sign and another line.

2 mole KClO3

1 mole KClO3

122.6 g KClO3

96.2 g KClO3

3 mole O2

Repeat Step 2: Draw a multiplication sign and another line.

1 mole KClO3

122.6 g KClO3

96.2 g KClO3

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

g KClO3

96.2 g KClO3

The balanced reaction shows the ratio of O2 to KClO3 is 3:2.

Repeat Step 5: Use the corresponding conversion factor to write in the values.

2 mole KClO3

1 mole KClO3

122.6 g KClO3

96.2 g KClO3

3 mole O2

2KClO3 → 2KCl + 3O2

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole O2

2 mole KClO3

3 mole O2

1 mole KClO3

122.6 g KClO3

96.2 g KClO3

32.0 g O2

1 mole of O2 has a mas of 32.0 grams.

Step 1: Put what you start with over 1.

96.2 g KClO3

Step 5: Use the corresponding conversion factor to write in the values.

One mole of KClO3 has a mass of 122.6 grams.

1 mole KClO3

122.6 g KClO3

96.2 g KClO3

Step 2: Draw a multiplication sign and another line.

96.2 g KClO3

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

61.5 g CO

1 Mole = 22.4 L of gas at STP

61.5 g CO

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

1 Mole = 6.02 x 1023 particles

Use balanced equation

particles A

particles B

61.5 g CO

g CO

28.0 g CO

g CO

28.0 g N2

1 mole N2

44.8 L N2

22.4 L N2

1 mole N2

2KClO3 → 2KCl + 3O2

2H2 + O2 → 2H2O

2KClO3 → 2KCl + 3O2

How many liters of O2 (at standard temperatuure and pressure) are produced when 44.8 grams of KClO3 decompose?

= 12.3 L O2

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

1 mole O2

2 mole KClO3

3 mole O2

1 mole KClO3

122.6 g KClO3

44.8 g KClO3

22.4 L O2

Repeat Step 4: Write the units of your next stop in the numerator.

mole O2

2 mole KClO3

3 mole O2

1 mole KClO3

122.6 g KClO3

44.8 g KClO3

L O2

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole O2

2 mole KClO3

3 mole O2

1 mole KClO3

122.6 g KClO3

44.8 g KClO3

22.4 L O2

1 mole of any gas at STP has a volume of 22.4 L.

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

1 mole KClO3

122.6 g KClO3

44.8 g KClO3

mole KClO3

Step 5: Use the corresponding conversion factor to write in the values.

One mole of KClO3 has a mass of 122.6 grams.

1 mole KClO3

122.6 g KClO3

44.8 g KClO3

Repeat Step 4: Write the units of your next stop in the numerator.

mole KClO3

1 mole KClO3

122.6 g KClO3

44.8 g KClO3

mole O2

2 mole KClO3

3 mole O2

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

1 mole KClO3

122.6 g KClO3

44.8 g KClO3

Repeat Step 2: Draw a multiplication sign and another line.

1 mole KClO3

122.6 g KClO3

44.8 g KClO3

Step 4: Write the units of your next stop in the numerator.

mole KClO3

g KClO3

44.8 g KClO3

Step 1: Put what you start with over 1.

44.8 g KClO3

The balanced reaction shows the ratio of O2 to KClO3 is 3:2.

Repeat Step 5: Use the corresponding conversion factor to write in the values.

2 mole KClO3

1 mole KClO3

122.6 g KClO3

44.8 g KClO3

3 mole O2

2KClO3 → 2KCl + 3O2

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

g KClO3

44.8 g KClO3

Repeat Step 2: Draw a multiplication sign and another line.

2 mole KClO3

1 mole KClO3

122.6 g KClO3

44.8 g KClO3

3 mole O2

Step 2: Draw a multiplication sign and another line.

44.8 g KClO3

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

1 Mole = 22.4 L of gas at STP

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

1 Mole = 6.02 x 1023 particles

Use balanced equation

particles A

particles B

2KClO3 → 2KCl + 3O2

How many molecules of O2 are produced when 215 grams of KClO3 decompose?

= 1.58 x 1024 molecules O2

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

1 mole O2

2 mole KClO3

3 mole O2

1 mole KClO3

122.6 g KClO3

215 g KClO3

6.02 x 1023 molecules O2

Repeat Step 4: Write the units of your next stop in the numerator.

mole O2

2 mole KClO3

3 mole O2

1 mole KClO3

122.6 g KClO3

215 g KClO3

molecules O2

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

1 mole KClO3

122.6 g KClO3

215 g KClO3

mole KClO3

mole O2

2 mole KClO3

3 mole O2

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

1 mole KClO3

122.6 g KClO3

215 g KClO3

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole O2

2 mole KClO3

3 mole O2

1 mole KClO3

122.6 g KClO3

215 g KClO3

6.02 x 1023 molecules O2

1 mole of O2 conntains 6.02 x 1023 molecules.

Repeat Step 4: Write the units of your next stop in the numerator.

mole KClO3

1 mole KClO3

122.6 g KClO3

215 g KClO3

mole O2

The balanced reaction shows the ratio of O2 to KClO3 is 3:2.

Repeat Step 5: Use the corresponding conversion factor to write in the values.

2 mole KClO3

1 mole KClO3

122.6 g KClO3

215 g KClO3

3 mole O2

2KClO3 → 2KCl + 3O2

Repeat Step 2: Draw a multiplication sign and another line.

1 mole KClO3

122.6 g KClO3

215 g KClO3

Step 4: Write the units of your next stop in the numerator.

mole KClO3

g KClO3

215 g KClO3

Repeat Step 2: Draw a multiplication sign and another line.

2 mole KClO3

1 mole KClO3

122.6 g KClO3

215 g KClO3

3 mole O2

Step 5: Use the corresponding conversion factor to write in the values.

One mole of KClO3 has a mass of 122.6 grams.

1 mole KClO3

122.6 g KClO3

215 g KClO3

Step 1: Put what you start with over 1.

215 g KClO3

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

g KClO3

215 g KClO3

Step 2: Draw a multiplication sign and another line.

215 g KClO3

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

1 Mole = 22.4 L of gas at STP

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

1 Mole = 6.02 x 1023 particles

Use balanced equation

particles A

particles B

22.4 L O2

2KClO3 → 2KCl + 3O2

How many moles of O2 are produced when 625 grams of KClO3 decompose?

Repeat Step 4: Write the units of your next stop in the numerator.

mole KClO3

1 mole KClO3

122.6 g KClO3

625 g KClO3

mole O2

= 7.65 moles O2

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

2 mole KClO3

3 mole O2

1 mole KClO3

122.6 g KClO3

625 g KClO3

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

1 mole KClO3

122.6 g KClO3

625 g KClO3

mole KClO3

Step 4: Write the units of your next stop in the numerator.

mole KClO3

g KClO3

625 g KClO3

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

g KClO3

625 g KClO3

Repeat Step 2: Draw a multiplication sign and another line.

1 mole KClO3

122.6 g KClO3

625 g KClO3

The balanced reaction shows the ratio of O2 to KClO3 is 3:2.

Repeat Step 5: Use the corresponding conversion factor to write in the values.

2 mole KClO3

1 mole KClO3

122.6 g KClO3

625 g KClO3

3 mole O2

2KClO3 → 2KCl + 3O2

Step 1: Put what you start with over 1.

625 g KClO3

Step 5: Use the corresponding conversion factor to write in the values.

One mole of KClO3 has a mass of 122.6 grams.

1 mole KClO3

122.6 g KClO3

625 g KClO3

Step 2: Draw a multiplication sign and another line.

625 g KClO3

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

1 Mole = 22.4 L of gas at STP

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

1 Mole = 6.02 x 1023 particles

Use balanced equation

particles A

particles B

22.4 L O2

2H2 + O2 → 2H2O

How many moles of O2 are needed for the combustion of 128 L of H2 at standard temperature and pressure?

Repeat Step 4: Write the units of your next stop in the numerator.

mole H2

1 mole H2

22.4 L H2

128 H2

mole O2

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

1 mole H2

22.4 L H2

128 L H2

mole H2

The balanced reaction shows the ratio of O2 to H2 is 1:2.

Repeat Step 5: Use the corresponding conversion factor to write in the values.

2 mole H2

1 mole H2

22.4 L H2

128 H2

1 mole O2

2H2 + 1O2 → 2H2O

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

L H2

128 L H2

Step 4: Write the units of your next stop in the numerator.

mole H2

L H2

128 L H2

= 2.86 moles O2

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

2 mole H2

1 mole O2

1 mole H2

22.4 L H2

128 L H2

Step 1: Put what you start with over 1.

128 L H2

Repeat Step 2: Draw a multiplication sign and another line.

1 mole H2

22.4 L H2

128 L H2

Step 5: Use the corresponding conversion factor to write in the values.

One mole of any gas at STP has a volume of 22.4 L

1 mole H2

22.4 L H2

128 L H2

Step 2: Draw a multiplication sign and another line.

128 L H2

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

1 Mole = 22.4 L of gas at STP

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

1 Mole = 6.02 x 1023 particles

Use balanced equation

particles A

particles B

22.4 L O2

2H2 + O2 → 2H2O

How many grams of H2O are produced when 60.0 L of O2 at standard temperature and pressure completely react with H2?

= 96.4 g H2O

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

1 mole H2O

1 mole O2

2 mole H2O

1 mole O2

22.4 L O2

60.0 L O2

18.0 g H2O

Repeat Step 4: Write the units of your next stop in the numerator.

mole H2O

1 mole O2

2 moles H2O

1 mole O2

22.4 L O2

60.0 L O2

g H2O

mole H2O

1 mole O2

2 moles H20

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

1 mole O2

22.4 L O2

60.0 L O2

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

1 mole O2

22.4 L O2

60.0 L O2

mole O2

Repeat Step 4: Write the units of your next stop in the numerator.

mole O2

1 mole O2

22.4 L O2

60.0 L O2

mole H2O

Step 1: Put what you start with over 1.

60.0 L O2

The balanced reaction shows the ratio of H2O to O2 is 2:1.

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole O2

22.4 L O2

60.0 L O2

2 moles H2O

2H2 + 1O2 → 2H2O

Step 4: Write the units of your next stop in the numerator.

mole O2

L O2

60.0 L O2

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

L O2

60.0 L O2

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole H2O

1 mole O2

2 moles H2O

1 mole O2

22.4 L O2

60.0 L O2

18.0 g H2O

1 mole of H2O has a mass of 18.0 grams.

Repeat Step 2: Draw a multiplication sign and another line.

1 mole O2

22.4 L O2

60.0 L O2

2 moles H20

Repeat Step 2: Draw a multiplication sign and another line.

1 mole O2

22.4 L O2

60.0 L O2

Step 5: Use the corresponding conversion factor to write in the values.

The volume of 1 mole of any gas at STP is 22.4 L.

1 mole O2

22.4 L O2

60.0 L O2

Step 2: Draw a multiplication sign and another line.

60.0 L O2

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

1 Mole = 22.4 L of gas at STP

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

1 Mole = 6.02 x 1023 particles

Use balanced equation

particles A

particles B

2H2 + O2 → 2H2O

How many molecules of H2O are produced when 30.0 L of O2 at standard temperature and pressure completely react with H2?

= 1.61 x 1024 molecules H2O

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

1 mole H2O

1 mole O2

2 mole H2O

1 mole O2

22.4 L O2

30.0 L O2

6.02 x 1023 molecules H2O

Repeat Step 4: Write the units of your next stop in the numerator.

1 mole O2

2 moles H2O

1 mole O2

22.4 L O2

30.0 L O2

mole H2O

molecules H2O

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

1 mole O2

22.4 L O2

30.0 L O2

mole O2

Repeat Step 4: Write the units of your next stop in the numerator.

mole O2

1 mole O2

22.4 L O2

30.0 L O2

mole H2O

Repeat Step 2: Draw a multiplication sign and another line.

1 mole O2

22.4 L O2

30.0 L O2

2 moles H20

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole O2

2 moles H2O

1 mole O2

22.4 L O2

30.0 L O2

1 mole of H2O contains 6.02 x 1023 molecules.

1 mole H2O

6.02 x 1023 molecules H2O

mole H2O

1 mole O2

2 moles H20

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

1 mole O2

22.4 L O2

30.0 L O2

Step 4: Write the units of your next stop in the numerator.

mole O2

L O2

30.0 L O2

The balanced reaction shows the ratio of H2O to O2 is 2:1.

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole O2

22.4 L O2

30.0 L O2

2 moles H2O

2H2 + 1O2 → 2H2O

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

L O2

30.0 L O2

Repeat Step 2: Draw a multiplication sign and another line.

1 mole O2

22.4 L O2

30.0 L O2

Step 2: Draw a multiplication sign and another line.

30.0 L O2

Step 1: Put what you start with over 1.

30.0 L O2

Step 5: Use the corresponding conversion factor to write in the values.

The volume of 1 mole of any gas at STP is 22.4 L.

1 mole O2

22.4 L O2

30.0 L O2

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

1 Mole = 22.4 L of gas at STP

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

1 Mole = 6.02 x 1023 particles

Use balanced equation

particles A

particles B

2H2 + 1O2 → 2H2O

mole H2

2C8H18 + 25O2 → 16CO2 + 18H2O

How many moles of CO2 are produced as a result of the combustion of 5.80 x 1024 molecules of C8H18?

Step 5: Use the corresponding conversion factor to write in the values.

One mole of C8H18 = 6.02 x 1023 molecules of C8H18.

5.80 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

= 77.1 moles CO2

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

16 moles CO2

2 moles C8H18

5.80 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

5.80 x 1024 molecules C8H18

molecules C8H18

Repeat Step 4: Write the units of your next stop in the numerator.

moles CO2

moles C8H18

5.80 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

moles C8H18

5.80 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

The balanced reaction shows the ratio of CO2 to C8H18 is 16:2.

Repeat Step 5: Use the corresponding conversion factor to write in the values.

16 moles CO2

2 moles C8H18

5.80 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 1: Put what you start with over 1.

5.80 x 1024 molecules C8H18

Step 2: Draw a multiplication sign and another line.

5.80 x 1024 molecules C8H18

Repeat Step 2: Draw a multiplication sign and another line.

5.80 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

Step 4: Write the units of your next stop in the numerator.

5.80 x 1024 molecules C8H18

molecules C8H18

mole C8H18

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

1 Mole = 22.4 L of gas at STP

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

3 mole O2

128 L H2

1 mole H2

1 Mole = 6.02 x 1023 particles

2 mole H2

22.4 L H2

Use balanced equation

particles A

particles B

22.4 L O2

2C8H18 + 25O2 → 16CO2 + 18H2O

How many grams of CO2 are produced as a result of the combustion of 2.60 x 1024 molecules of C8H18?

= 1520 g CO2

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

1 mole CO2

16 moles CO2

2 moles C8H18

2.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

44.0 g CO2

Step 5: Use the corresponding conversion factor to write in the values.

2.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

One mole of C8H18 = 6.02 x 1023 molecules of C8H18.

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

2.60 x 1024 molecules C8H18

molecules C8H18

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

moles C8H18

2.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

Repeat Step 4: Write the units of your next stop in the numerator.

moles CO2

moles C8H18

2.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

Repeat Step 4: Write the units of your next stop in the numerator.

mole CO2

16 moles CO2

2 moles C8H18

2.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

g CO2

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole of CO2 has a mass of 44.0 grams.

1 mole CO2

16 moles CO2

2 moles C8H18

2.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

44.0 g CO2

16 moles CO2

2 moles C8H18

2.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

Repeat Step 2: Draw a multiplication sign and another line.

Step 1: Put what you start with over 1.

2.60 x 1024 molecules C8H18

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

mole CO2

16 moles CO2

2 moles C8H18

2.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

Step 4: Write the units of your next stop in the numerator.

2.60 x 1024 molecules C8H18

molecules C8H18

mole C8H18

Step 2: Draw a multiplication sign and another line.

2.60 x 1024 molecules C8H18

Repeat Step 5: Use the corresponding conversion factor to write in the values.

16 moles CO2

2 moles C8H18

2.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

2C8H18 + 25O2 → 16CO2 + 18H2O

The balanced reaction shows the ratio of CO2 to C8H18 is 16:2.

Repeat Step 2: Draw a multiplication sign and another line.

2.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

1 Mole = 22.4 L of gas at STP

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

1 Mole = 6.02 x 1023 particles

Use balanced equation

particles A

particles B

2C8H18 + 25O2 → 16CO2 + 18H2O

How many liters (at STP) of CO2 are produced as a result of the combustion of 7.60 x 1024 molecules of C8H18?

Step 5: Use the corresponding conversion factor to write in the values.

7.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

One mole of C8H18 = 6.02 x 1023 molecules of C8H18.

1 mole CO2

16 moles CO2

2 moles C8H18

7.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

22.4 L CO2

= 2260 L CO2

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

7.60 x 1024 molecules C8H18

molecules C8H18

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

moles C8H18

7.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

Repeat Step 4: Write the units of your next stop in the numerator.

moles CO2

moles C8H18

7.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole of any gas at STP has a volume of 22.4 L.

1 mole CO2

16 moles CO2

2 moles C8H18

7.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

22.4 L CO2

16 moles CO2

2 moles C8H18

7.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

Repeat Step 2: Draw a multiplication sign and another line.

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

mole CO2

16 moles CO2

2 moles C8H18

7.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

Repeat Step 4: Write the units of your next stop in the numerator.

mole CO2

16 moles CO2

2 moles C8H18

7.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

L CO2

Step 1: Put what you start with over 1.

7.60 x 1024 molecules C8H18

Step 2: Draw a multiplication sign and another line.

7.60 x 1024 molecules C8H18

Repeat Step 2: Draw a multiplication sign and another line.

7.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

Repeat Step 5: Use the corresponding conversion factor to write in the values.

16 moles CO2

2 moles C8H18

7.60 x 1024 molecules C8H18

6.02 x 1023 molecules C8H18

1 mole C8H18

2C8H18 + 25O2 → 16CO2 + 18H2O

The balanced reaction shows the ratio of CO2 to C8H18 is 16:2.

Step 4: Write the units of your next stop in the numerator.

7.60 x 1024 molecules C8H18

molecules C8H18

mole C8H18

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

1 Mole = 22.4 L of gas at STP

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

1 Mole = 6.02 x 1023 particles

Use balanced equation

particles A

particles B

2KClO3 → 2KCl + 3O2

How many liters of O2 (at standard temperatuure and pressure) are produced when 25.5 moles of KClO3 decompose?

= 857 L O2

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

1 mole O2

22.4 L O2

3 moles O2

2 moles KClO3

25.5 moles KClO3

Repeat Step 4: Write the units of your next stop in the numerator.

mole O2

L O2

3 moles O2

2 moles KClO3

25.5 moles KClO3

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole O2

22.4 L O2

The volume of 1 mole of any gas at STP is 22.4 L.

3 moles O2

2 moles KClO3

25.5 moles KClO3

mole O2

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

3 moles O2

2 moles KClO3

25.5 moles KClO3

Step 1: Put what you start with over 1.

25.5 moles KClO3

Step 5: Use the corresponding conversion factor to write in the values.

The balanced reaction shows the ratio of O2 to KClO3 is 3:2.

2KClO3 → 2KCl + 3O2

3 moles O2

2 moles KClO3

25.5 moles KClO3

Repeat Step 2: Draw a multiplication sign and another line.

3 moles O2

2 moles KClO3

25.5 moles KClO3

Step 2: Draw a multiplication sign and another line.

25.5 moles KClO3

Step 4: Write the units of your next stop in the numerator.

moles O2

moles KClO3

25.5 moles KClO3

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

moles KClO3

25.5 moles KClO3

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

1 Mole = 22.4 L of gas at STP

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

1 Mole = 6.02 x 1023 particles

Use balanced equation

particles A

particles B

44.8 g KClO3

3 mole O2

1 mole KClO3

122.6 g KClO3

2 mole KClO3

One mole of KClO3 has a mass of 122.6 grams.

2KClO3 → 2KCl + 3O2

How many molecules of O2 are produced when 40.8 moles of KClO3 decompose?

= 3.68 x 1025 molecules O2

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

3 moles O2

2 moles KClO3

40.8 moles KClO3

1 mole O2

6.02 x 1023 molecules O2

Repeat Step 4: Write the units of your next stop in the numerator.

mole O2

molecules O2

3 moles O2

2 moles KClO3

40.8 moles KClO3

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole O2

6.02 x 1023 molecules O2

1 mole of O2 contains 6.02 x 1023 molecules.

3 moles O2

2 moles KClO3

40.8 moles KClO3

mole O2

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

3 moles O2

2 moles KClO3

40.8 moles KClO3

Repeat Step 2: Draw a multiplication sign and another line.

3 moles O2

2 moles KClO3

40.8 moles KClO3

Step 1: Put what you start with over 1.

40.8 moles KClO3

Step 5: Use the corresponding conversion factor to write in the values.

3 moles O2

2 moles KClO3

40.8 moles KClO3

The balanced reaction shows the ratio of O2 to KClO3 is 3:2.

2KClO3 → 2KCl + 3O2

Step 2: Draw a multiplication sign and another line.

40.8 moles KClO3

moles O2

Step 4: Write the units of your next stop in the numerator.

moles KClO3

40.8 moles KClO3

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

moles KClO3

40.8 moles KClO3

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

1 Mole = 22.4 L of gas at STP

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

1 Mole = 6.02 x 1023 particles

Use balanced equation

particles A

particles B

44.8 g KClO3

3 mole O2

1 mole KClO3

122.6 g KClO3

2 mole KClO3

One mole of KClO3 has a mass of 122.6 grams.

2KClO3 → 2KCl + 3O2

How many grams of KCl are produced when 36.5 moles of KClO3 decompose?

= 2720 g KCl

Step 6: Cancel out units that are in both the numerator and denominator. Then, multiply the top numbers and divide by the bottom numbers to get your answer!

1 mole KCl

74.5 g KCl

2 moles KCl

2 moles KClO3

36.5 moles KClO3

Repeat Step 4: Write the units of your next stop in the numerator.

mole KCl

g KCl

2 moles KCl

2 moles KClO3

36.5 moles KClO3

mole KCl

Repeat Step 3: Write the units you want to cancel out in the denominator of this new fraction.

2 moles KCl

2 moles KClO3

36.5 moles KClO3

Step 1: Put what you start with over 1.

36.5 moles KClO3

Repeat Step 2: Draw a multiplication sign and another line.

2 moles KCl

2 moles KClO3

36.5 moles KClO3

Repeat Step 5: Use the corresponding conversion factor to write in the values.

1 mole KCl

74.5 g KCl

1 mole of KCl has a mass of 74.5 grams.

2 moles KCl

2 moles KClO3

36.5 moles KClO3

Step 3: Write the units you want to cancel out in the denominator of this new fraction.

moles KClO3

36.5 moles KClO3

Step 2: Draw a multiplication sign and another line.

36.5 moles KClO3

Step 5: Use the corresponding conversion factor to write in the values.

2 moles KCl

2 moles KClO3

36.5 moles KClO3

The balanced reaction shows the ratio of KCl to KClO3 is 2:2.

2KClO3 → 2KCl + 3O2

moles KCl

Step 4: Write the units of your next stop in the numerator.

moles KClO3

36.5 moles KClO3

volume of gas A (liters)

volume of gas B (liters)

Use balanced equation

1 Mole = 22.4 L of gas at STP

mass A (grams)

moles A

mass B (grams)

moles B

Use balanced equation

Use Molar Mass

1 Mole = 6.02 x 1023 particles

Use balanced equation

particles A

particles B

44.8 g KClO3

3 mole O2

1 mole KClO3

122.6 g KClO3

2 mole KClO3

One mole of KClO3 has a mass of 122.6 grams.

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You need the orange element on every page. It will also display errors (in Spanish) This element shows either as "baraja1" or as "set1". Copy and paste to create as many as you have cards. Group the element with your cards IN THE ORDER that you want them to appear. Go to a differrent slide and return to see the updated numbering. The numbering changes with every times you group/ungroup or duplicate elements. Optional Elements: Group the "contador" with any text. The text will be automatically replaced to display how many rounds have been played. Group the "Terminado" element with text/object that you want to appear once all cards have been used. Group the "Reiniciar" element with a button (e.g. an image or text) . When the button is pressed, the game is reset. Group the "Ocultar" with a text/object that you want to disappear once the game starts, e.g. the instructions Choose one type of these buttons to decide how cards will be drawn:

true

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display a new element, hiding the rest

display a new element, keeping the rest visible

display a random element with animation.

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This page allows for up to 5 decks of cards. See next page if you need more.For each set of cards, use a different set of "baraja/set" elements. Each cards of deck 1 need to be grouped with a "baraja/set 1" button, each card of deck 2 need to be grouped with a "baraja/set 2" button etc. Also use a different set of set of buttons from the white box for each set (Ocultar 1 for the first set, Ocultar 2 for the second set etc.)

Group the elements from the orange box with buttons that will apply to ALL sets of cards (e.g. it draw the next card in all sets at the same time or it will reset all cards at the same time)

Elements for up to 10 sets If you need more sets, use the codes on the next 2 pages. Change the numbers in bolt to create more elements. You can also change the colour names in the codes to give it different colours to be able to tell them apart easier. Once the code is changed, go to "insert", then "other" and paste it in the box to create the element.

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