The Study of The Study of StoichiometryStoichiometryThe Study of The Study of
StoichiometryStoichiometry
I. I. Stoichiometric Stoichiometric CalculationsCalculations
A. Proportional A. Proportional RelationshipsRelationshipsA. Proportional A. Proportional RelationshipsRelationships
How many eggs are needed to make 12 dozen cookies?
3/4 c. brown sugar1 tsp vanilla extract2 eggs2 c. chocolate chipsMakes 5 dozen cookies.
2 1/4 c. flour1 tsp. baking soda1 tsp. salt1 c. butter3/4 c. sugar
12 doz. 2 eggs
5 doz.= 5 eggs
Ratio of eggs to cookies
A. Proportional A. Proportional RelationshipsRelationshipsA. Proportional A. Proportional RelationshipsRelationships
StoichiometryStoichiometry• calculating amounts of reactants &
products using mole ratios
Mole RatioMole Ratio• indicated by coefficients in a
balanced equation
2 Mg + O2 Mg + O22 2 MgO 2 MgO
A. Proportional A. Proportional RelationshipsRelationshipsA. Proportional A. Proportional RelationshipsRelationships
Mole Ratio examples:Mole Ratio examples:
2 H2 H22 + O + O22 2 H 2 H22OO2 mol H2 mol H22 : 1 mol O : 1 mol O22 2 mol H 2 mol H22OO
CHCH4 4 + 2O+ 2O22 CO CO22 + 2H + 2H22OO1 mol CH1 mol CH4 4 : 2 mol O: 2 mol O22 1 mol CO1 mol CO2 2 : 2 mol H: 2 mol H22OO
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
Core step in all stoichiometry problems!!
• Mole ratio - moles moles
4. Check answer.
C. Stoichiometry C. Stoichiometry ProblemsProblemsC. Stoichiometry C. 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
C. Stoichiometry C. Stoichiometry ProblemsProblemsC. Stoichiometry C. Stoichiometry ProblemsProblems
How many grams of KCl will be formed from 2.5 mol KClO3?
2.5 mol KClO3
2 molKCl
2 molKClO3
= 186.3 g KCl
2KClO3 2KCl + 3O2
74.55 gKCl
1 molKCl
2.5 mol ? g
C. Stoichiometry C. Stoichiometry ProblemsProblemsC. Stoichiometry C. 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
C. Stoichiometry C. Stoichiometry ProblemsProblemsC. Stoichiometry C. Stoichiometry ProblemsProblems
1. Try on your ownHow many grams of calcium carbonate are required to prepare 50.0 g of calcium oxide?
CaCO3 CaO + CO2
C. Stoichiometry C. Stoichiometry ProblemsProblemsC. Stoichiometry C. Stoichiometry ProblemsProblems
C. Stoichiometry C. Stoichiometry ProblemsProblemsC. Stoichiometry C. Stoichiometry ProblemsProblems
II. Gas StoichiometryII. Gas Stoichiometry
Ch. 5 - StoichiometryCh. 5 - StoichiometryCh. 5 - StoichiometryCh. 5 - Stoichiometry
1 mol of a gas=22.4 Lat STP
A. Molar Volume at STPA. Molar Volume at STPA. Molar Volume at STPA. Molar Volume at STP
Standard Temperature & Pressure0°C and 1 atm
A. Molar Volume at STPA. Molar Volume at STPA. Molar Volume at STPA. 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)
B. Gas Stoichiometry B. Gas Stoichiometry ProblemProblemB. Gas Stoichiometry B. Gas Stoichiometry ProblemProblem
How many grams of CaCO3 are req’d to
produce 9.00 L of CO2 at STP?
9.00 LCO2
1 molCO2
22.4 L CO2
= 40.2 g CaCO3
CaCO3 CaO + CO2
1 molCaCO3
1 molCO2
100.09g CaCO3
1 molCaCO3
? g 9.00 L
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 2.1 mol HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?
Zn + 2HCl ZnCl2 + H2 79.1 g ? L2.1 mol
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 ? L2.1 mol
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
22.4L H2
1 molH2
2.1 mol HCl
1 mol H2
2 mol HCl
= 23.52 L H2
Zn + 2HCl ZnCl2 + H2 79.1 g ? L2.1 mol
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
Zn: 27.1 L H2 HCl: 23.5 L H2
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 23.52 L H2
left over zinc
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
35.6 g of zinc react with 1.9 mol HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?
Zn + 2HCl ZnCl2 + H2 35.6 g ? L1.9 mol
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
35.6g Zn
1 molZn
65.39g Zn
= 12.2 L H2
1 molH2
1 molZn
22.4 LH2
1 molH2
Zn + 2HCl ZnCl2 + H2 35.6 g ? L1.9 mol
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
22.4L H2
1 molH2
1.9 mol HCl
1 mol H2
2 mol HCl
= 21.28 L H2
Zn + 2HCl ZnCl2 + H2 35.6 g ? L1.9 mol
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
Zn: 12.2 L H2 HCl: 21.28 L H2
Limiting reactant: Zn
Excess reactant: HCl
Product Formed: 12.2 L H2
left over HCl
B. Percent YieldB. Percent YieldB. Percent YieldB. Percent Yield
100yield ltheoretica
yield actualyield %
calculated on paper
measured in lab
B. Percent YieldB. Percent YieldB. Percent YieldB. 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
B. Percent YieldB. Percent YieldB. Percent YieldB. 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:
B. Percent YieldB. Percent YieldB. Percent YieldB. 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
2 HCl + Zn H2 + ZnCl2130.7 g Zn was able to produce
twice its mass of Zinc Chloride in the chemical reaction above. What
is the percent yield of ZnCl2?
=130.7 1
65.39
g Zn Mol Zn
g Zn272.4 g ZnCl2
g ZnCl2136.28
mol ZnCl2
1
Percent Yield
1 mol Zn
mol ZnCl21
= Actual
Calculated
261.4 g ZnCl2
272.4 g ZnCl2
X 100= 95.9% Yield
HCl + NaOH NaCl + H NaCl + H2O
100 g HCl reacts with 100 g NaOH in the chemical reaction above.
What is the limiting reactant and how many grams of NaCl will be
produced by the reaction?
=100 1
36.45
g HCl mol HCl
g HCl
160.3 g NaCl
g NaCl58.44
mol NaCl1
=100 1
40.0
g NaOH
mol NaOHg NaOH
146.1 g NaCl
mol NaCl1
mol NaOH
1 mol NaCl
1
g NaCl58.44
Limiting Reactant
1 mol HCl
mol NaCl1