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STOICHIOMETRY
Mole-Mole
Mole-Mass
Mass-Mass
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STOICHIOMETRYSTOICHIOMETRYSTOICHIOMETRYSTOICHIOMETRY
- the study of the - the study of the quantitative quantitative aspects of aspects of chemical chemical reactions.reactions.
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STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS
STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS
Stoichiometricfactor
Molesreactant
Moles product
Coef. WantCoef. given
Mole -Mole Problems
Moles of the
Given Substance
Moles of the
Unknown Substance
Conversion Factor:Coefficients from the balanced equation
Example Problem 2HCl + Fe = FeCl2 + H2
How many mole of HCl are need to react with 6 mole of Fe?
6 mol Fe X 2HCl1 Fe
= 12 mol HCl
Mole -Mass Problems (2 step)
Moles of the
Given Substance
Mass of the given Substance
Conversion Factor:Molar mass of the given
X
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STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS
STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS
Mass reactant
StoichiometricfactorMoles
reactantMoles product
Mass product
Molar massgiven
Molar massUnknown
Example Problem 2HCl + Fe = FeCl2 + H2
How many grams of HCl are need to react with 6 mole of Fe?
6 mol Fe X 2HCl1 Fe
= 12 mol HCl
12 mol HCl X36 g HCl1 mol
= 432g
Mass -Mass Problems (3 step)
Mass of the Given Substance
Mass of the
Unknown Substance
Conversion Factors:Molar mass of each, and
Coefficients from the balanced equation
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STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS
STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS
Mass reactant
StoichiometricfactorMoles
reactantMoles product
Mass product
Molar massgiven
Molar massUnknown
= 1 mol Fe
2HCl + Fe = FeCl2 + H2
How many grams of HCl are need to react with 56 g of Fe?
1 mol Fe X2HCl1 Fe
= 2 mol HCl
2 mol HCl X36 g HCl1 mol
= 72g HCl
1mol 56 g Fe
56 g of Fe X
1212
PROBLEM: If 454 g of NHPROBLEM: If 454 g of NH44NONO33 decomposes, how much Ndecomposes, how much N22O and O and HH22O are formed? What is the O are formed? What is the theoretical yield of products?theoretical yield of products?
PROBLEM: If 454 g of NHPROBLEM: If 454 g of NH44NONO33 decomposes, how much Ndecomposes, how much N22O and O and HH22O are formed? What is the O are formed? What is the theoretical yield of products?theoretical yield of products?
STEP 1STEP 1
Write the balanced Write the balanced chemical equationchemical equation
NHNH44NONO33 ---> --->
NN22O + 2 HO + 2 H22OO
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454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO 454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO
STEP 2 STEP 2 Convert mass reactant Convert mass reactant (454 g) --> moles(454 g) --> moles
454 g • 1 mol
80.04 g = 5.68 mol NH4NO3
STEP 3 STEP 3 Convert moles reactant Convert moles reactant (5.68 mol) --> moles product(5.68 mol) --> moles product
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STEP 3 STEP 3 Convert moles reactant --> moles productConvert moles reactant --> moles product
Relate moles NHRelate moles NH44NONO33 to moles product to moles product
expected. expected.
1 mol NH1 mol NH44NONO33 --> 2 mol H --> 2 mol H22OO
Express this relation as the Express this relation as the STOICHIOMETRICSTOICHIOMETRIC
FACTORFACTOR..2 mol H2O produced1 mol NH4NO3 used
2 mol H2O produced1 mol NH4NO3 used
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454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO 454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO
= 11.4 mol H= 11.4 mol H22O producedO produced
5.68 mol NH4NO3 • 2 mol H2O produced1 mol NH4NO3 used
STEP 3 STEP 3 Convert moles reactant (5.68 Convert moles reactant (5.68 mol) --> moles productmol) --> moles product
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454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO 454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO
11.4 mol H2O • 18.02 g1 mol
= 204 g H2O
STEP 4 STEP 4 Convert moles product Convert moles product (11.4 mol) --> mass product(11.4 mol) --> mass product
Called the Called the THEORETICAL THEORETICAL YIELDYIELD
ALWAYS FOLLOW THESE STEPS IN ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY SOLVING STOICHIOMETRY
PROBLEMS!PROBLEMS!
ALWAYS FOLLOW THESE STEPS IN ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY SOLVING STOICHIOMETRY
PROBLEMS!PROBLEMS!
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454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO 454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO
STEP 5 STEP 5 How much NHow much N22O is formed?O is formed?
Total mass of reactants = Total mass of reactants =
total mass of productstotal mass of products
454 g NH454 g NH44NONO33 = ___ g N = ___ g N22O + 204 g HO + 204 g H22OO
mass of Nmass of N22O = 250. gO = 250. g
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454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO 454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO
STEP 6 STEP 6 Calculate the Calculate the percent percent
yieldyieldIf you isolated only 131 g of NIf you isolated only 131 g of N22O, what is O, what is
the percent yield?the percent yield?
This compares the This compares the theoreticaltheoretical (250. g) (250. g)
and and actualactual (131 g) yields. (131 g) yields.
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454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO 454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22OO
% yield = actual yield
theoretical yield • 100%
STEP 6 STEP 6 Calculate the percent yieldCalculate the percent yield
% yield = 131 g250. g
• 100% = 52.4%
2020
PROBLEM: Using 5.00 g of PROBLEM: Using 5.00 g of HH22OO22, what mass of O, what mass of O22 and of and of HH22O can be obtained?O can be obtained?
PROBLEM: Using 5.00 g of PROBLEM: Using 5.00 g of HH22OO22, what mass of O, what mass of O22 and of and of HH22O can be obtained?O can be obtained?
2 H2 H22OO22(liq) ---> 2 H(liq) ---> 2 H22O(g) + OO(g) + O22(g)(g)
Reaction is catalyzed by MnOReaction is catalyzed by MnO22
Step 1: moles of HStep 1: moles of H22OO22
Step 2: use STOICHIOMETRIC FACTOR Step 2: use STOICHIOMETRIC FACTOR to calculate moles of Oto calculate moles of O22
Step 3: mass of OStep 3: mass of O22
