Stoichiometry

9.1 Calculating Quantities in Reactions

• Determine mole ratios from a balanced chemical equation

• Explain why mole ratios are central to solving stoichiometry problems

• Solve stoichiometry problems involving:– Mass using molar mass– Volume using denisty– Particles using Avogadro’s number

Ham Sandwich

How many sandwiches could you make from 24 slices of bread?

How many slices of• lettuce would you need?• tomato?• ham?• cheese?

This process models the calculations in this chapter.

Equations are like recipes.

2C8H18 + 25O2 → 16CO2 + 18H2O

• Coefficients can be read as ratios of particles or of moles.If 2 mols of C8H18 reacts completely how many moles of CO2 would be produced?

Stoichoimetry = • the proportional relationship between 2 or

more substances during a chemical reaction• quantitative analysis of the outcomes of a

reaction• Predict amount of product you could make

from starting amounts of reactant

The Mole Ratio is the Key

• In stoichiometry problems, the unit that bridges the gap between one substance and another is the mole.

• You can use the coefficients in conversion factors called mole ratios

Sample Problem A

N2 + 3H2 → 2NH3

• 3 mols of hydrogen are needed to prepare 2 moles of ammonia. 3 mol H2 = 2 mol NH3

PROBLEM: How many moles of hydrogen are needed to prepare 312 moles of ammonia?

312 mol NH3 x 3 mol H2 = ? mol H2

2 mol NH3

= 468 mol H2

needed

Practice Prob. P. 3041. Calculate the amounts requested if 1.34 mol H2O2

completely react according to the following equation2H2O2 → 2H2O + O2

a. mols of oxygen formedGiven: 1.34 mol H2O2 reacts

2 mol H2O2 = 1 mol O2

1.34 mol H2O2 x 1 mol O2 = 0.670 mol O2 produced

2 mol H2O2

Practice Prob. P. 3041. Calculate the amounts requested if 1.34 mol H2O2

completely react according to the following equation2H2O2 → 2H2O + O2

b. mols of water formedGiven: 1.34 mol H2O2 reacts

2 mol H2O2 = 2 mol H2O

1.34 mol H2O2 x 2 mol H2O = 1.34 mol H2O produced

2 mol H2O2

2. Calculate the amounts requested if 3.30 mol Fe2O3 completely react according to the following equationFe2O3 + 2Al → 2Fe + Al2O3 Thermite Reaction

a. mols of aluminum neededGiven: 3.30 mol Fe2O3 reacts

1 mol Fe2O3 = 2 mol Al needed

3.30 mol Fe2O3 x 2 mol Al = 6.60 mol Al needed

1 mol Fe2O3

2. Calculate the amounts requested if 3.30 mol Fe2O3 completely react according to the following equationFe2O3 + 2Al → 2Fe + Al2O3

b. mols of iron formedGiven: 3.30 mol Fe2O3 reacts

1 mol Fe2O3 = 2 mol Fe formed

3.30 mol Fe2O3 x 2 mol Fe = 6.60 mol Fe formed

1 mol Fe2O3

2. Calculate the amounts requested if 3.30 mol Fe2O3 completely react according to the following equationFe2O3 + 2Al → 2Fe + Al2O3

c. mols of aluminum oxide formedGiven: 3.30 mol Fe2O3 reacts

1 mol Fe2O3 = 1 mol Al2O3 formed

3.30 mol Fe2O3 x 1 mol Al2O3 = 3.30 mol Al2O3 formed

1 mol Fe2O3

Application/Homework

Worksheet “Stoichiometry: Mole-Mole Problems” – front only

Stoichiometry: Mass-Mass Problems

• You must convert to moles using molar mass of known

• Then use mole ratio to find moles unknown• Convert back to mass using molar mass of

unknown

Solving Mass-Mass Problems

Mass Unknown

1 mol___ grams

Periodic Table

Mol unknownMol known

BalancedChemicalEquation

___ grams1 mol

Periodic Table

Sample Problem B p. 307What mass of NH3 can be made from 1221 g H2 and excess N2?

N2 + 3H2 → 2 NH3

1221 g H2 X 1 mol H2 x 2 mol NH3 x 17.04g NH3

2.02 g 3 mol H2 1 mol NH3

GRAMS MOLAR MOLAR MOLARKNOWN MASS RATIO MASS

KNOWN UNKNOWN= 6867 g NH3 made

Practice p. 307#1-4Use the equation below to answer #1-4 Fe2O3 + 2Al → 2Fe + Al2O3

1. How many grams Al needed to completely react with 135 grams Fe2O3

135 g Fe2O3 X 1 mol Fe2O3 x 2 mol Al x 26.98 g Al

159.7 g Fe2O3 1 mol Fe2O3 1 mol Al

GRAMS MOLAR MOLAR MOLARKNOWN MASS RATIO MASS

KNOWN UNKNOWN

= 45.6 g Al needed

Practice p. 307#1-4Use the equation below to answer #1-4 Fe2O3 + 2Al → 2Fe + Al2O3

2. How many grams Al2O3 can form when 23.6g Al react with excess Fe2O3?

23.6 g Al X 1 mol Al x 1 mol Al2O3 x 101.96 g Al2O3

26.98 g Al 2 mol Al 1 mol Al2O3

= 44.6 g Al2O3 formed

Practice p. 307#1-4Use the equation below to answer #1-4 Fe2O3 + 2Al → 2Fe + Al2O3

3. How many grams of Fe2O3 react with excess Al to make 475 g Fe?

475 g Fe X 1 mol Fe x 1 mol Fe2O3 x 159.7 g Fe2O3

55.85 g Fe 2 mol Fe 1 mol Fe2O3

= 679 g Fe2O3 react

Practice p. 307#1-4Use the equation below to answer #1-4 Fe2O3 + 2Al → 2Fe + Al2O3

4. How many grams of Fe will form when 97.6 g Al2O3 form?

97.6 g Al2O3 X 1 mol Al2O3 x 2 mol Fe x 55.85 g Fe

101.96 g Al2O3 1 mol Al2O3 1 mol Fe

= 107 g Fe formed

Homework

Worksheet “Stoichiometry: Mass-Mass Problems” – back of mol-mol worksheet

Lab: Mass Relations in Chemical Reactions (Baking soda & HCl reaction)

Stoichiometry:Volume-Volume Problems

• Liquid amounts often measured in volumes. You might use density and molar mass for these problems p. 308 Toolkit

• Density unit g/mL means ____g = 1 mL• Gases 22.41 L = 1 mol of any gas• Basics the same: Change to moles, use mole

ratio, and then change to desired units.

DensityDensityDensity

Solving Volume-Volume Problems

Volume Known

Volume Unknown

Sample Problem C p. 309What volume of H3PO4 forms when 56 mL POCl3 completely react? (density of POCl3 = 1.67 g/mL; density of H3PO4 = 1.83 g/mL)POCl3(l) + 3H2O(l) → H3PO4 (l) + 3HCl (g)

56mL POCl3 x 1.67 g POCl3 x 1 mol POCl3 x 1 mol H3PO4

1 mL 153.32g 1 mol POCl3

X 98.00 g H3PO4 x 1 mL H3PO4 = 33 mL H3PO4

1 mol 1.83 g

Practice p. 309 # 1-4C5H12 (l) → C5H8(l) + 2H2(g)

Densities: C5H12 = 0.620 g/mL

C5H8 = 0.681 g/mL

H2 = 0.0899 g/L

= 0.0899 g/1000mL = 8.99 x 10-5 g/mLCalculateMolar Masses: C5H12 = 72.17 g/mol

C5H8 = 68.13 g/mol

H2 = 2.02 g/mol

1. How many mL of C5H8 can be made from 366mL C5H12?

366mL C5H12 x 0.620 g x 1 mol C5H12 x 1 mol C5H8

1 mL 72.17 g 1 mol C5H12

X 68.13 g x 1 mL 1 mol C5H8 0.681 g

= 315 mL C5H8

2. How many liters of H2 can form when 4.53 x 103 mL C5H8 form?

4.53 x 103 mLC5H8 x 0.681 g x 1 mol C5H8 x 2 mol H2

1 mL 68.13 g 1 mol C5H8

X 2.02 g x 1 L 1 mol H2 0.0899 g

= 2030 L H2

3. How many mL of C5H12 are needed to make97.3 mL of C5H8?

97.3 mL C5H8 x 0.681 g x 1 mol C5H8 x 1 mol C5H12

1 mL 68.13 g 1 mol C5H8

X 72.17 g x 1 mL 1 mol C5H12 0.620 g

= 113 mL C5H12

4. How many milliliters of H2 can be made from 1.98 x 103 mL C5H12?

1.98 x 103 mL C5H12 x 0.620 g x 1 mol C5H12 x 2 mol H2

1 mL 72.17 g 1 mol C5H12

X 2.02 g x 1 mL 1 mol H2 0.0000899 g

= 7.64 x 105 mL H2

Homework

• WS “Stoichiometry: Volume-Volume Problems” front only

• WS “Stoichiometry: MixedProblems” back – classwork

9.1 QUIZ will be on __________________

9.2 Limiting Reactants and Percentage Yield

In this section you will:

• Identify the limiting reactant for a reaction and use it to calculate theoretical yield.

• Perform calculations involving percentage yield.

Limiting Reactant• To drive a car you need gasoline and oxygen from the air.• When the gas runs out, you can’t go any farther even though there is

still plenty of oxygen.• The gasoline limits the distance you can travel because it runs out first.

Page 312/313 Figures 3&4 Which of the starting supplies limited the number of mums that could be made? Which items where in excess/left over?

Limiting reactant = substance that controls the quantity of product that can form in a chemical reaction; runs out first

Excess reactant = substance that is not used up completely in a reaction.

Identify the Limiting Reactant from Lab Data• Calculate the amount of product that each could form.• Whichever reactant would produce

the least amount of product is the limiting reactant.

Theoretical yield = maximum amount of product that can be made if everything about the reaction works out perfectly; determined by the limiting reactant.

• Whenever a problem gives you quantities of 2 or more reactants, you must

1. Determine the limiting reactant (makes less product)

2. Use the limiting reactant quantity to determine the theoretical yield.

Sample E p. 314

Identify the limiting reactant and the theoretical yield of phosphorous acid, H3PO3 if 225g of PCl3 is mixed with 123 g of H2O.

PCl3 + 3H2O → H3PO3 + 3HCl

• Determine how many g of H3PO4 that each reactant could make.• Need to calculate molar masses to use as conversion factors.

225g PCl3 x 1mol PCl3 X 1mol H3PO4 x 82.00g H3PO4

137.32g 1 mol PCl3 1 mol

= 134 g H3PO4

123g H2O x 1mol PCl3 X 1mol H3PO4 x 82.00g H3PO4

18.02g 3 mol H2O 1 mol

= 187 g H3PO4

•PCl3 is the limiting reactant because 134 g is less than 187 g.•The theoretical yield is 134 grams of H3PO3.

Practice p. 314 #1-3PCl3 + 3H2O → H3PO3 + 3HClIdentify the limiting reactant and the theoretical yield (in grams) of HCl for each pair

1. 3.00 mol PCl3 and 3.00 mol H2O

3.00 mol PCl3 x 3 mol HCl = 9.00 mol HCl

1 mol PCl3

3.00 mol H2Ox 3 mol HCl = 3.00 mol HCl

3 mol H2O

H2O is limiting reactant.

3.00 mol HCl x 36.51 g = 109 grams HCl

1 mol theoretical yield (made)

Practice p. 314 #1-3PCl3 + 3H2O → H3PO3 + 3HClIdentify the limiting reactant and the theoretical yield (in grams) of HCl for each pair

2. 75.0 g PCl3 and 75.0 g H2O

75.0g PCl3 x 1 mol PCl3 x 3 mol HCl = 1.65 mol HCl

136.5 g 1 mol PCl3

75.0 g H2O x 1 mol H2O x 3 mol HCl = 4.16 mol HCl

18.02 3 mol H2O

PCl3 is limiting reactant.

1.65 mol HCl x 36.51 g = 60.2 grams HCl

1 mol theoretical yield (made)

Practice p. 314 #1-3PCl3 + 3H2O → H3PO3 + 3HClIdentify the limiting reactant and the theoretical yield (in grams) of HCl for each pair

3. 1.00 mol PCl3 and 50.0 g H2O

1.00 mol PCl3 x 3 mol HCl = 3.00 mol HCl

1 mol PCl3

50.0 g H2O x 1 mol H2O x 3 mol HCl = 2.77 mol H3PO3

18.02 3 mol H2O

H2O is limiting reactant.

2.77 mol HCl x 36.51 g = 101 grams HCl

1 mol theoretical yield (made)

Limiting Reactants & Industry

• The most expensive chemicals are chosen as the limiting reactants

• Less expense reactants can be used in excess to ensure all of the expensive chemicals are completely used up (none wasted).

Actual Yield and Percentage Yield

• Although equations tell you what should happen, they cannot always tell you what will happen in real life/in the lab.

• Some reactions do not make all of the product predicted by the theoretical yield.ACTUAL YIELD = the mass of product actually formed, measured in lab, often less than expected (theoretical).

Examples of things that could reduce yield:

• Incomplete distillation/purification needed to separate product from a mixture.

• Side reactions that can use up reactants without making desired products

PERCENTAGE YIELD = ratio relating the actual to the theoretical yield; describes how efficient the reaction was.

Percentage Yield = actual x 100 theoretical

SHOULD ALWAYS BE LESS THAN 100%. If not, you switched the actual & theoretical in the formula!!!

Sample Problem F p. 317N2 + 3 H2 → 2NH3

Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N2 are mixed with 9.0 g H2 and 16.1 g NH3 form.

14.0g N2 x 1 mol N2 x 2 mol NH3 x 17.04 g NH3 = 17.0g NH3

28.02 g 1 mol N2 1 mol NH3

9 .0g H2 x 1 mol N2 x 2 mol NH3 x 17.04 g NH3 = 51 g NH3

2.02 g 3 mol H2 1 mol NH3

N2 makes the smaller amount and is the limiting reactant.

Sample Problem F p. 317N2 + 3 H2 → 2NH3

Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N2 are mixed with 9.0 g H2 and 16.1 g NH3 form.

• N2 makes the smaller amount and is the limiting reactant.

• The theoretical amount made is 17.0g NH3.

• The actual amount made is 16.1 g NH3

PERCENTAGE YIELD = 16.1 g x 100 = 94.7% 17.0g

Practice p. 317 #1N2 + 3 H2 → 2NH3

1. Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N2 , 3.15g H2 and actual is 14.5 g NH3.

14.0g N2 x 1 mol N2 x 2 mol NH3 x 17.04 g NH3 = 17.0g NH3

28.02 g 1 mol N2 1 mol NH3

3.15g H2 x 1 mol N2 x 2 mol NH3 x 17.04 g NH3 = 17.7 g NH3

2.02 g 3 mol H2 1 mol NH3

N2 makes the smaller amount and is the limiting reactant.

N2 + 3 H2 → 2NH3

Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N2 are mixed with 9.0 g H2 and 16.1 g NH3 form.

• N2 makes the smaller amount and is the limiting reactant.

• The theoretical amount made is 17.0g NH3.

• The actual amount made is 14.5 g NH3

PERCENTAGE YIELD = 14.5 g x 100 = 85.3% 17.0g

Homework

Section 9.2 Review p. 319 # 1,3,4,5,6,8,10,12,14

Quiz on 9.2 will be on __________________.