stoichiometric calculations start your book problems now!! stoichiometry
TRANSCRIPT
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:
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