REACTIONS & STOICHIOMETRY

UNIT 6

Overview Reactions

Write formula/word equationsBalance EquationsIdentify TypesPredict ProductsWrite Net Ionic Equations

StoichiometryConversionsLimiting & Excess ReagentPercent Yield

Chemical Reactions Process in which one or more pure substances are

converted into one or more different pure substances

Reactants: Zn + I2 Product: Zn I2

Indications of a Reaction Temperature Change Color Change Production of gas Formation of a precipitate Production of light

Chemical Equations4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)

Reactants react to produce products

The letters (s), (g), (l), and (aq) are the physical states of compounds.“aq” represents aqueous meaning dissolved in water (solution)

The numbers in the front are called coefficients.

Subscripts represent the number of each atom in a compound

(Reactants) (Products)

Chemical ReactionsSymbol Meaning

+ used to separate one reactant or product from another

→ used to separate the reactants from the products - it is pronounced "yields" or "produces" when the equation is read

↔ used when the reaction can proceed in both directions - this is called an equilibrium arrow and will be used later in the course

↑ an alternative way of representing a substance in a gaseous state

↓ an alternative way of representing a substance in a solid state

∆ indicates that heat is applied to make the reaction proceed

Diatomic Elements Elements that cannot exist by

themselves (always occur in pairs)Bromine (Br2)Iodine (I2)Nitrogen (N2)Chlorine (Cl2)Hydrogen (H2)Oxygen (O2)Fluorine (F2)

BrINClHOF!

Writing Equations Practice1. When lithium hydroxide pellets are added to a

solution of sulfuric acid, lithium sulfate and water are formed.

2 LiOH(s) + H2SO4(aq) Li2SO4(aq) + 2 H2O(l)

2. When crystalline C6H12O6 is burned in oxygen, carbon dioxide and water vapor are formed.

C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g)

Balancing Equations Law of Conservation of Mass

Matter cannot be destroyed (atoms of reactants must equal products)

Balance equations to get same number of each atom on the left and right in an equation

2 Hg atoms, 2 O atoms 2 Hg atoms, 2 O atoms2HgO(s) ---> 2 Hg(l) + O2(g)

Balancing Equations___ Al(s) + ___ Br2(l) ---> ___ Al2Br6(s)

__C3H8(g) + __ O2(g) __ CO2(g) + __ H2O(g)

__ B4H10(g) + __ O2(g) __ B2O3(g) + __ H2O(g)

6 Types of Reactions Synthesis (combination) Decomposition Single Replacement (displacement) Double Replacement (precipitation) Combustion Acid-Base Neutralization

Synthesis (Combination) Reactions Two or more substances combine to form a

new compound.

A + X AX

Synthesis of: Binary compounds H2 + O2 H2O Metal carbonates CaO + CO2 CaCO3 Metal hydroxides CaO + H2O Ca(OH)2 Metal chlorates KCl + O2 KClO3

Oxyacids CO2 + H2O H2CO3

Decomposition Reactions A single compound breaks down into two

or more simpler substances

AX A + X

Decomposition of: Binary compounds H2O H2 + O2

Metal carbonates CaCO3 CaO + CO2

Metal hydroxides Ca(OH)2 CaO + H2O Metal chlorates KClO3 KCl + O2

Oxyacids H2CO3 CO2 + H2O

Single Replacement (displacement) Reactions One element replaces another in a

reactionMetals replace metalsNonmetals replace nonmetals

A + BX AX + B

BX + Y BY + X

Activity Series Decide whether or not one element will

replace another Metals can replace other metals provided

that they are above the metal that they are trying to replace

If the metal is not above what it is trying to replace, the result is “no reaction”

Double Replacement (Precipitation) Reactions Two elements or ions “switch partners”

AX + BY AY + BX One of the compounds

formed is usually a precipitate, an insoluble gas that bubbles out of solution, or a molecular compound, usually water.

Solubility Solubility – ability to dissolve

In a double replacement (precipitate) reaction, one of the products must be insoluble in water and form a precipitatePrecipitate – insoluble solid formed by a reaction in solutionIf both products are soluble the result is “no reaction”

Solubility rules help you determine whether or not a compound will form a precipitate or remain an aqueous solution

Solubility RulesSoluble Ionic Compounds Except with:Alkali metals, NH4

+

NO3-, C2H3O2

-, ClO3-, ClO4

- (no exceptions)

Cl-, Br-, I- Ag+, Hg2+2, Pb+2

SO4-2 Sr+2, Ba+2, Ca+2, Ag+, Pb+2, Hg2

+2

Insoluble Ionic Compounds Except with:CO3

-2, PO4-3, SiO3

-2, O-2, SO3-2, CrO4

-2 NH4+, alkali metals

S-2 NH4+, alkali metals

Ca+2, Sr+2, Ba+2, Mg+2 (group 2)

OH- NH4+, alkali metals,

(Ca+2, Ba+2, Sr+2 are slightly soluble)

Combustion Reactions A substance combines with oxygen,

releasing a large amount of energy in the form of light and heat.Produces a flameFuel + oxygen produces carbon dioxide and

water vapor

CxHx + O2 CO2 + H2O

Acid-Base Neutralization Reactions When the solution of an acid and solution of a

base are mixed Products have no characteristics of either the

acid or the base Acid + Base (metal hydroxide) salt + water

Salt comes from cation of base and anion of acid

HY + XOH XY + H2O

Chemical Equations Molecular Equation – shows complete chemical formulas of

reactants and products

Pb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq)

Complete Ionic Equation – All soluble electrolytes shown as ions

Pb+2(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) PbI2(s) + 2K+(aq) + 2NO3

-(aq)

Net Ionic Equation – shows only the ions and molecules directly involved in the equation

Pb+2(aq) + 2I-(aq) PbI2(s)

Writing Complete Ionic Equations

1. Start with a balanced molecular equation. 2. Break all soluble strong electrolytes

(compounds with (aq) beside them) into their ions.

indicate the correct formula and charge of each ion

indicate the correct number of each ion write (aq) after each ion

3. Bring down all compounds with (s), (l), or (g) unchanged.

Writing Complete Ionic Equations

Example:

2Na3PO4(aq) + 3CaCl2(aq) 6NaCl(aq) + Ca3(PO4)2(s)

Becomes…

6Na+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Cl-(aq) 6Na+(aq) + 6Cl-

(aq) + Ca3(PO4)2(s)

Spectator Ions Appear in identical forms among both

the reactants and products of a complete ionic equation

When writing net ionic equations they cancel each other out

Pb+2(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) PbI2(s) + 2K+(aq) + 2NO3

-(aq)

Writing Net Ionic Equations Cancel out spectator ions from complete ionic

equation then write what’s left

6Na+(aq) + 2PO43-(aq) + 3Ca2+(aq) + 6Cl-(aq) 6Na+(aq) + 6Cl-

(aq) + Ca3(PO4)2(s)

Becomes…

2PO43-(aq) + 3Ca2+(aq) Ca3(PO4)2(s)

PracticeWrite complete ionic and net ionic equations for

the following:

1. 3(NH4)2CO3(aq) + 2Al(NO3)3(aq) 6NH4NO3(aq) + Al2(CO3)3(s)

2. 2NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2H2O(l)

3. Zn(s) + CuSO4(aq) --> ZnSO4(aq) + Cu(s)

Answers1. Complete Ionic Equation:

6NH4+(aq) + 3CO3

2-(aq) + 2Al3+(aq) + 6NO3-(aq) 6NH4

+(aq) + 6NO3-(aq) + Al2(CO3)3(s)

Net Ionic Equation: 2 Al3+(aq) + 3 CO32-(aq) Al2(CO3)3(s)

2. Complete Ionic Equation:2Na+(aq) + 2OH-(aq) + 2H+(aq) + SO4

2-(aq) 2Na+(aq) + SO42-(aq) + 2H2O(l)

Net Ionic Equation: OH-(aq) + H+(aq) H2O(l)*Note: simplify net ionic equations if possible

3. Complete Ionic Equation:Zn(s) + Cu2+(aq) + SO4

2-(aq) Zn2+(aq) + SO42-(aq) + Cu(s)

Net Ionic Equation: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

Stoichiometry

The study of the quantitative aspects of chemical reactions.

Mole Ratio Conversion factor that relates amount in

moles of any two substances involved in a chemical reaction

2 Al2O3(l) 4 Al(s) + 3 O2(g)

Mole ratio Al2O3 to O2 = 2:3Mole ratio Al to Al2O3 = 4:2 or 2:1Mole ratio Al to O2 = 4:3

Stoichiometry Problems Solved just like conversions! You must start with a balanced chemical

equation Types:

Mole MoleMass MassMass Mole or Mole Mass

Mole Mole

How many moles of O2 are produced from 3.5 moles of Al2O3?

3.5 mol Al2O3 × = 5.25 mol O2

*Use mole ratio to convert between moles!

2 Al2O3(l) 4 Al(s) + 3 O2(g)

3 mol O2

2 mol Al2O3

Mass Mass

How many grams of Al are produced from 4.56 grams of Al2O3?

Molar Mass Al2O3 = 101.96 g/mol Molar Mass Al = 26.98 g/mol

4.56 g Al2O3 × × × = 2.41 g Al

1 mol Al2O3

101.96 g Al2O3

4 mol Al2 mol Al2O3

26.98 g Al1 mol Al

2 Al2O3(l) 4 Al(s) + 3 O2(g)

Limiting/Excess ReactantRecipe makes 10 pancakes

3 eggs2 cups bisquik1 cup milk1 cup chocolate chips

What is the most amount of pancakes that I can make with 6 eggs and 5 cups of milk?

What is the most amount of pancakes that I can make with 3 cups of chocolate chips and 8 cups of milk?

What “limits” how many pancakes I

can make and what will be left over?

Limiting/Excess Reactant The limiting reactant is the reactant that is

consumed first, limiting the amounts of products formed.

The excess reactant is the reactant that is leftover after the reaction has gone to completion.

Reactants Products2 NO(g) + O2 (g) 2 NO2(g)

Limiting reactant = ___________Excess reactant = ____________

Limiting/Excess Reactant

Calculating Limiting/Excess Reagent

Given 12.4 grams of NO and 9.40 grams of O2, which is the limiting and which is the excess reagent?

2 NO(g) + O2 (g) 2 NO2(g)

1 mol NO30.01 g NO

2 mol NO2

2 mol NO46.01 g NO2

1 mol NO2

12.4 g NO × × × = 19.01 g NO2

1 mol O2

16.00 g O2

2 mol NO2

1 mol O2

46.01 g NO2

1 mol NO2

9.40 g O2 × × × = 54.06 g NO2

Calculating Limiting/Excess Reagent

NO limits the amount of NO2 that is madeLimiting reagent = NO

O2 will be leftover once the reaction is completeExcess reactant = O2

2 NO(g) + O2 (g) 2 NO2(g)1 mol NO

30.01 g NO2 mol NO2

2 mol NO46.01 g NO2

1 mol NO2

12.4 g NO × × × = 19.01 g NO2

1 mol O2

32.00 g O2

2 mol NO2

1 mol O2

46.01 g NO2

1 mol NO2

9.40 g O2 × × × = 27.03 g NO2

Calculating Limiting/Excess Reagent

How much O2 will be in excess once the reaction is complete?

2 NO(g) + O2 (g) 2 NO2(g)

1 mol NO30.01 g NO

1 mol O2

2 mol NO32.00 g O2

1 mol O212.4 g NO × × × = 6.61 g O2

6.61 grams of O2 will be used in the reaction. You have 9.40 grams to start with.

9.40 – 6.61 = 2.79 grams O2 in excess (leftover)

Limiting/Excess Reactant If the equation has 2 or more products, when

determining the limiting/excess reactants, simply pick one of the products and convert both reactants to that product.

You MUST use the same product for both.

Percent Yield

Percentage comparing how much product was actually produced compared to what should have been produced.

Calculate theoretical yield using stoichiometry. If you know how much of each reactant you start

out with, use stoichiometry to calculate how much of the given product you should produce.

Actual YieldTheoretical Yield

× 100

Percent Yield

An experiment was performed combining using 3.4 g of AgNO3 and an unlimited supply of KCl. If the experiment yielded 2.7 g of AgCl, what is the percent yield of the experiment?

AgNO3(aq) + KCl(aq) → AgCl(s) + KNO3(aq)

1 mol AgNO3

169.88 g AgNO3

1 mol AgCl1 mol AgNO3

143.32 g AgCl1 mol AgCl

3.4 g AgNO3 × × × = 2.9 g AgCl

Percent Yield = × 100 = 93%2.72.9