Want to create interactive content? It’s easy in Genially!

Get started free

Chemistry Final Project

Start designing with a free template

Discover more than 1500 professional designs like these:

Blackboard Presentation

Genial Storytale Presentation

Historical Presentation

Psychedelic Presentation

Memories Presentation

Animated Chalkboard Presentation

Chalkboard Presentation

Transcript

Chemistry Unit 4: Chemical Reactions

By: Cayden Lamar

START

Index

Identifying Reaction Types

Predicting Products

Balancing Equations

Synthesis

Combustion

Decomposition

Single Replacement

Double Replacement

Practice Problems

Additional Resources

Before balancing equations you need to know the:

Subscripts and coefficients

Subscripts are how many atoms or ions are within a compound Coefficients are how many moles or units there are You cannot change the subscripts You can change the coefficients Why? This is because changing the subscripts changes the compound meanwhile changing the coefficients only changes the amount there are.

The steps for balancing equations are:

1. Identify all the elements in the equation. They should be the same for products and reactants 2. Using the subscripts and coefficients, count the total number of elements on each side of the equation and ask yourself "Is it balanced?" 3. Use coefficiants to balance the equation

How to Identify Reaction Types

It's important to note that some reactions may fall into multiple categories or involve a combination of different reaction types. There are other specialized reaction types such as combustion reactions (involving a substance reacting with oxygen and releasing energy). This is how you can identify which reaction you are doing based on the equation:

Synthesis

Combustion

Decomposition

Single Replacement

Double Replacement

Ways To Predict Products

Balancing the equation will also help you predict the number of reactants and products. Finally, consider the solubility rules. In double displacement reactions, consider the solubility rules to predict whether you need to replace an element or not.

Some ways you can predict the problem are by identifying the reaction type. It could be a synthesis reaction, decomposition reaction, displacement reaction, double displacement reaction, or combustion reaction, among others. This identification will provide a starting point for predicting the products. You must also consider the reactivity of elements. Elements have different reactivities. Use the activity series to determine if a displacement reaction is possible. For example, metals higher in the reactivity series can displace metals lower in the series from their compounds in a single replacement reaction.

Synthesis Reactions

A + B → AB Reactants are just two elements then Just combine the elements Metals first and criss-cross oxidation numbers (charges)

Description: Two reactants come together to form one product General Equation: A + B → AB Example: 2 Cu + O2→ 2 CuO

Key: M = Metal; NM = Nonmetal A. Formation of binary compound: A + B → AB B. Metal oxide and water: MO + H2O → base C. Nonmetal oxide and water: (NM)O + H2O → acid

Base = Metal(OH) Acid = H NonMetal C is the special one! You can use subscripts from the reactants

Examples:

Decomposition Reactions

Diatomic elements are elements that exist in pairs Iodine (I2) I Bromine (Br2) Bring Chlorine (Cl2) Cookies Fluorine (F2) For Oxygen (O2) Our Nitrogen (N2) New Hydrogen (H2) Home The diatomic elements are the ones you need to look for as they can effect the product.

Description: 1 reactant breaks apart to form 2 or more products General Equation: AB → A + B Example: 2 H2CO3→ H2O + CO2

Key: M = Metal; NM = Nonmetal a. Binary compounds: AB → A + B b. Metallic carbonates: MCO3 → MO + CO c. Metallic hydrogen carbonates: MHCO3 → MCO3 (s) + H2O (l) + CO2 (g) d. Metallic hydroxides: MOH → MO + H2O e. Metallic chlorates: MClO3 → MCl + O2 f. Oxyacids decompose to nonmetal oxides and water: Oxyacid → (NM)O + H2O The decomposition reaction is always one big element no more no less.

Examples

Description:A hydrocarbon reacts with O2 to produce CO2 and H2O General Equation: hydrocarbon + O2 → CO2 + H2O Example: C3H8 + O2→ CO2 + H2O You will always have CO2 and H2O as the products!

Combustion Reactions

Examples

Products are always the same. When balancing a combustion equation you should always balance the oxygen last. And if you are stuck balancing you should try doubling coefficients.

Some single replacement reactions DO NOT occur because some elements are not as active as others. There is a list called the Activity Series on your Chemistry Reference Packet.

Single Replacement

Description: Element replaces another in a compound (1 element ends up "single")General Equation: AB+C→ A + BC Example: 2Fe+3H2O → 3H2 +Fe2O3

Key: M = Metal; NM = Nonmetal A. Metal-Metal replacement: A + BC → AC + B B. Active metal replaces H from water: M + H2O → MOH + H2 C. Active metal replaces H from acid: M + HX → MX + H2 d. Halide-Halide replacement: D + BC → BD + C

Examples

A higher (more active) element on the activity series replaces lower element. If the element by itself is lower on the list, the reaction will NOT occur.

Double Replacement

Description: Ions trade partners, a precipitate forms General Equation: AB + CD → AD + CB Example: CaCl2+2AgNO3→ Ca(NO3)2+2AgCl

Acid-Base Neutralization

Positive Ions swap placesWhen looking for an Acid-Base equation one product will always be H2O Regular Equation

H2SO3 + NaOH → The positive ions change places.

Na(OH) + FeCl3 →The positive ions change place.

H2SO3 + NaOH → Na(SO3) + H(OH)Find the chages to criss-cross

Na(OH) + FeCl3 → Fe(OH) + NaClFind the charges to criss-cross

H2SO3 + NaOH → Na1+(SO3)2- + H1+(OH)1- Then we SWAP, DROP, and CHOP

Na(OH) + FeCl3 → Fe3+(OH)1- + Na1+Cl1- Then we need to SWAP, DROP, and CHOP Na(OH) + FeCl3 → Fe(OH)3 + NaCl Now we need to balance! 3Na(OH) +FeCl3 → Fe(OH)3 + 3NaCl

H2SO3 + NaOH → Na2(SO3) + H2O Now we need to balance!

H2SO3 + 2NaOH → Na2(SO3) + 2H2O

A Real Worlds Connection

One example of a double replacement reaction that connects to a real-world phenomenon is the reaction between sodium chloride (NaCl) and silver nitrate (AgNO₃) to form silver chloride (AgCl) and sodium nitrate (NaNO₃): NaCl + AgNO₃ → AgCl + NaNO₃

This reaction is observed in photography. In traditional film photography, silver halides (such as silver chloride) are used in the film emulsion. When exposed to light, a chemical reaction occurs, causing a reduction in the silver halides and the formation of metallic silver. This reaction allows the image to be captured in the film. The reaction between sodium chloride and silver nitrate represents a simplified version of the chemical process involved in the development of photographic film.

NaCl + AgNO₃ → AgCl + NaNO₃

Examples

Synthesis Practice Problems

16

_K2O + _H2O →2K(OH)

K2O + H2O →K1- (OH)1- → K(OH)

K- 1X→2H- 1X→2 O- 1X→2

K- 2H- 2 O- 2

H- 2 S- 1 O- 4

SO3 + H2O → H2 (SO4)4 → H2SO4

_SO3 + _H2O → _H2SO4

H- 2 S- 1 O- 4

23

30

Decomposition Practice Problems

Because the Bromine is a diatomic element the charge for the product will be 2. This will leave the charges on both sides to be equal and leave teh equation balanced. So the final answer would be CaBr2 → Ca + Br2

CaBr2 → Ca + Br2

Ca- 1Br- 2

Ca- 1Br- 2

Li2CO3 → Li1+ O2- Li2O + CO2

Li2CO3 → Li2O + CO2Li-2 C- 1 O- 3

Li-2C- 1 O- 3

Combustion Practice Problems

_C8H8 + _10_O2 → _8_CO2+_4_H2O *Products are always CO2 +H2O*C- 8 H- 8 O- 2X →10

C- 1X→8H- 2X→8 O- 3X→20

_2C2H6 + _7O2 → _4CO2 + _6H2OC- 2X→4 H- 6X→12 O- 2X→14

C- 1X→4H- 2X→12 O- 3X→14

Single Replacement Practice Problems

K + NaCl → K is higher on the activity seriesK1+ Na1- K + NaCl → KCl + Na

The answer to Sn + FeCl3 is No reaction because on the activity series Sn is not higher then Fe. Meaning that Sn does not have a higher activity then Fe. This means that Sn would not be able to replace Fe having no reaction.

Double Replacement Practice Problems

AB + CD → AD + CB

CaCl2 + NaOH → Ca- 1 Cl- 2 Na- 1X → 2 O-1X → 2 H- 1X → 2

Ca- 1 Cl- 1X → 2 Na- 1X → 2 O- 2 H- 2

CaCl2 + _2NaOH → _2NaCl + Ca(OH)2

H3PO4 + CuOH →Cu I (OH)-1 1x=1 1X1=1 X=1

H- 4X → 6P- 1 O- 5X → 7 Cu- 1X → 3

H- 2X → 6P- 1 O- 5 → 7 Cu- 3

H3PO4 + 3CuOH → Cu3PO4+3H2O

Additional Resources To Help you!

Eureka!

How to predict products for Synthesis Reactions Predicting Products for Decomposition Reactions Predicting Products of Single Replacement Reactions Predicting Products for Double Replacement Reactions How to Predict Products of Chemical Reactions | How to Pass Chemistry