I. Stoichiometric
Calculations
Topic 9 Topic 9 StoichiometryStoichiometry
StoichiometryStoichiometryStoichiometryStoichiometry
The study of quantitative relationships between amounts of reactants used and amount of products formed in a reaction.
A. Proportional A. Proportional RelationshipsRelationshipsA. Proportional A. Proportional RelationshipsRelationships
StoichiometryStoichiometry• mass relationships between
substances in a chemical reaction• based on the mole ratio
Mole RatioMole Ratio• indicated by coefficients in a
balanced equation
2 Mg + O2 Mg + O22 2 MgO 2 MgO
Mole Ratio: Ratio between the number of moles of any two substances in a balanced chemical equation.
2 Mg + O2 2 Mg + O2 2 MgO 2 MgO
B. Stoichiometry StepsB. Stoichiometry StepsB. Stoichiometry StepsB. Stoichiometry Steps
1. Write a balanced equation.2. Identify known & unknown.3. Line up conversion factors.
• Mole ratio - moles moles• Molar mass - moles grams• Molarity - moles liters soln• Molar volume - moles liters gas
Core step in all stoichiometry problems!!
• Mole ratio - moles moles
4. Check answer.
C. Molar Volume at STPC. Molar Volume at STPC. Molar Volume at STPC. Molar Volume at STP
Molar Mass(g/mol)
6.02 1023
particles/mol
MASSIN
GRAMSMOLES
NUMBEROF
PARTICLES
LITERSOF
SOLUTION
Molar Volume (22.4 L/mol)
LITERSOF GASAT STP
Molarity (mol/L)
Steps to solving Steps to solving StoichiometryStoichiometrySteps to solving Steps to solving StoichiometryStoichiometry
Convert known quantity to moles.Convert moles of known to moles
of unknown using mole ratios.Convert unknown moles to
wanted quantity.
D. Stoichiometry D. Stoichiometry ProblemsProblemsD. Stoichiometry D. Stoichiometry ProblemsProblems
How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?
9 mol O2 2 mol KClO3
3 mol O2
= 6 mol KClO3
2KClO3 2KCl + 3O2 ? mol 9 mol
How many grams of KClO3 are req’d to
produce 9.00 L of O2 at STP?
9.00 LO2
1 molO2
22.4 L O2
= 32.8 g KClO3
2 molKClO3
3 molO2
122.55g KClO3
1 molKClO3
? g 9.00 L
D. Stoichiometry D. Stoichiometry ProblemsProblemsD. Stoichiometry D. Stoichiometry ProblemsProblems
2KClO3 2KCl + 3O2
D. Stoichiometry D. Stoichiometry ProblemsProblemsD. Stoichiometry D. Stoichiometry ProblemsProblems
How many grams of silver will be formed from 12.0 g copper?
12.0g Cu
1 molCu
63.55g Cu
= 40.7 g Ag
Cu + 2AgNO3 2Ag + Cu(NO3)2
2 molAg
1 molCu
107.87g Ag
1 molAg
12.0 g ? g
63.55g Cu
1 molCu
D. Stoichiometry D. Stoichiometry ProblemsProblemsD. Stoichiometry D. Stoichiometry ProblemsProblems
How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3?
1.5L
.10 molAgNO3
1 L= 4.8 g
Cu
Cu + 2AgNO3 2Ag + Cu(NO3)2
1 molCu
2 molAgNO3
? g 1.5L0.10M
E. Percent YieldE. Percent YieldE. Percent YieldE. Percent Yield
100yield ltheoretica
yield actualyield %
calculated on paper
measured in lab
E. Percent YieldE. Percent YieldE. Percent YieldE. Percent Yield
When 45.8 g of K2CO3 react with excess
HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.
K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g
actual: 46.3 g
E. Percent YieldE. Percent YieldE. Percent YieldE. Percent Yield
45.8 gK2CO3
1 molK2CO3
138.21 gK2CO3
= 49.4g KCl
2 molKCl
1 molK2CO3
74.55g KCl
1 molKCl
K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g
actual: 46.3 g
Theoretical Yield:
E. Percent YieldE. Percent YieldE. Percent YieldE. Percent Yield
Theoretical Yield = 49.4 g KCl
% Yield =46.3 g
49.4 g 100 =93.7%
K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g 49.4 g
actual: 46.3 g
II. Stoichiometry in the Real World
Topic 9 Topic 9 StoichiometryStoichiometry
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
Available IngredientsAvailable Ingredients• 4 slices of bread• 1 jar of peanut butter• 1/2 jar of jelly
Limiting ReactantLimiting Reactant• bread
Excess ReactantsExcess Reactants• peanut butter and jelly
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
Limiting ReactantLimiting Reactant• used up in a reaction• determines the amount of product
Excess ReactantExcess Reactant• added to ensure that the other
reactant is completely used up• cheaper & easier to recycle
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
1. Write a balanced equation.
2. For each reactant, calculate the
amount of product formed.
3. Smaller answer indicates:
• limiting reactant
• amount of product
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?
Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L
2.5M
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
79.1g Zn
1 molZn
65.39g Zn
= 27.1 L H2
1 molH2
1 molZn
22.4 LH2
1 molH2
Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L
2.5M
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
22.4L H2
1 molH2
0.90L
2.5 molHCl
1 L= 25 L
H2
1 molH2
2 molHCl
Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L
2.5M
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
Zn: 27.1 L H2 HCl: 25 L H2
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 25 L H2
left over zinc