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Stoichiometric
Calculations
Start Your Book Problems NOW!!
Stoichiometric
Calculations
Start Your Book Problems NOW!!
StoichiometryStoichiometry
A. Proportional A. Proportional RelationshipsRelationshipsA. Proportional A. Proportional RelationshipsRelationships
I have 5 eggs. How many cookies can I make?
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
5 eggs 5 doz.
2 eggs= 12.5 dozen cookies
Ratio of eggs to cookies
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
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
How many grams of KClO3 are req’d to
produce 9.00 mol of oxygen gas?
9.00 molO2
= 735 g KClO3
2 molKClO3
3 molO2
122.55g KClO3
1 molKClO3
? g 9.00 mol
C. Stoichiometry C. Stoichiometry ProblemsProblemsC. Stoichiometry C. Stoichiometry ProblemsProblems
2KClO3 2KCl + 3O2
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
II. Stoichiometry in the Real World
II. Stoichiometry in the Real World
Stoichiometry Part IIStoichiometry Part II
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 2.25 mol of HCl. Identify the limiting and excess reactants. How many grams of hydrogen gas are formed.
Zn + 2HCl ZnCl2 + H2 79.1 g ? g2.25 mol
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
79.1g Zn
1 molZn
65.39g Zn
= 2.44 g H2
1 molH2
1 molZn
2.02 gH2
1 molH2
Zn + 2HCl ZnCl2 + H2 79.1 g ? g2.25 mol
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
2.02 gL H2
1 molH2
2.25 molHCl
= 2.27 g H2
1 molH2
2 molHCl
Zn + 2HCl ZnCl2 + H2 79.1 g ? g2.25 mol
A. Limiting ReactantsA. Limiting ReactantsA. Limiting ReactantsA. Limiting Reactants
Zn: 2.44 g H2 HCl: 2.27 g H2
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 2.27 g H2
left over zinc
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: