stoichiometric calculations stoichiometry – ch. 8
TRANSCRIPT
Stoichiometri
c Calculations
Stoichiometry – Ch. 8Stoichiometry – Ch. 8
Proportional Proportional RelationshipsRelationshipsProportional 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
Proportional Proportional RelationshipsRelationshipsProportional 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
Stoichiometry StepsStoichiometry StepsStoichiometry StepsStoichiometry 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.
Stoichiometry ProblemsStoichiometry ProblemsStoichiometry ProblemsStoichiometry Problems
If you have 45 kg of isoamyl alcohol and enough acetic acid to react with all of the isoamyl alcohol, what is the maximum number of kg of isoamyl acetate that can be made?
C5H11OH + CH3COOH CH3COOC5H11 + H20
Stoichiometry ProblemsStoichiometry ProblemsStoichiometry ProblemsStoichiometry Problems
Equation balanced?
C5H11OH + CH3COOH CH3COOC5H11 + H20
YEA!!!
Stoichiometry ProblemsStoichiometry ProblemsStoichiometry ProblemsStoichiometry Problems
What do you know?45 kg of C5H11OH? kg of isoamyl acetateMolar mass of C5H11OH = 88.17g/molMolar mass of CH3COOC5H11= 130.21
g/mol 1 mol C5H11OH : 1 mol CH3COOC5H11 1000 g = 1 kg
Stoichiometry ProblemsStoichiometry ProblemsStoichiometry ProblemsStoichiometry Problems
45 kg 1mol C5H11OH 1000 g
88.17 g C5H11OH 1kg
= 510.mol C5H11OH
510.mol C5H11OH 1 mol CH3COOC5H11 =
1 mol C5H11OH
510. mol CH3COOC5H11
Stoichiometry ProblemsStoichiometry ProblemsStoichiometry ProblemsStoichiometry Problems
510. mol CH3COOC5H11 130.21 g 1kg
1 mol CH3COOC5H11 1000g=66.4 kg isoamyl acetate
66 kg isoamyl acetate is the maximum amount that can be produced
Stoichiometry ProblemsStoichiometry ProblemsStoichiometry ProblemsStoichiometry Problems
Magnesium burns in oxygen to produce magnesium oxide. How much magnesium will burn in the presence of 189 ml of oxygen? The density of oxygen is 1.429 g/L.
2Mg + O2 → 2MgO
Stoichiometry ProblemsStoichiometry ProblemsStoichiometry ProblemsStoichiometry Problems
189 ml O2
D = 1.429 g/L
1 mol O2 = 2 mol Mg
MM O2 = 32.00 g/ mol
MM Mg = 24.30 g/ mol
Stoichiometry ProblemsStoichiometry ProblemsStoichiometry ProblemsStoichiometry Problems
189 ml O2 • 1.429 g • 1 L • 1 mol O2 1 L 1000 ml 32.00 g O2
= 8.44 X 10-3 mol O2
8.44 X 10-3 mol O2 • 2 mol Mg 1 mol O2
= 16.88 X 10-3 mol Mg
16.88 X 10-3 mol Mg • 24.30 g Mg = 1 mol Mg
0.410 g Mg
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting 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
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting 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
Limiting Reactants- StepsLimiting Reactants- StepsLimiting Reactants- StepsLimiting Reactants- Steps
1. Write a balanced equation.
2. Determine # of moles present for each reactant.
3. Mole ratios but this time not for what you are looking for, but for reactants.
Limiting Reactants- StepsLimiting Reactants- StepsLimiting Reactants- StepsLimiting Reactants- Steps
4. Use the limiting reactant to solve the problem as usual .
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
Example problem page 286
CO + H2 → CH3OH Not balanced!
CO+ 2H2 → CH3OH
1 C 1 C
1 O 1 O
4 H 4 H
BALANCED!!!!!!!!!!!
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
Looking for kg of CH3OH produced
152.5 kg CO
MM CO = 28.01 g/mol
24.5 kg H2 MM H2 = 2.02 g/mol
1mol CO = 2mol H2
MM CH3OH = 32.05g/mol
1mol CO = 1mol CH3OH
2mol H2 = 1mol CH3OH
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
Determine # of moles present for each reactant.
152.5 kg CO x 1mol CO x 1000g =
28.01 g CO 1 kg
5.444 x 103 mol CO
24.5 kg H2 x 1mol x 1000 g =
2.02 g H2 1 kg
1.213 x 104 mol H2
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
Mole ratios but this time not for product for reactants
1 mol CO = 2 mol H2
1.213 x 104 mol H2 x 1mol CO =
2 mol H2
6.065 x 103 mol CO
Need 6.065 X 103 mol CO
Have 5.444 X 103 mol CO
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
CO is the limiting reactant!!!!
Not enough to react with all of the H2
present.
Use the limiting reactant to solve the problem
Limiting ReactantsLimiting ReactantsLimiting ReactantsLimiting Reactants
5.444 X 103 mol CO 1 mol CH3OH
1 mol CO
32.05 g CH3OH 1 kg
1 mol CH3OH 1000 g
= 174.5 kg CH3OH
Percent YieldPercent YieldPercent YieldPercent Yield
100yield ltheoretica
yield actualyield %
calculated on paper
measured in lab
Percent YieldPercent YieldPercent YieldPercent 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
Percent YieldPercent YieldPercent YieldPercent 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:
Percent YieldPercent YieldPercent YieldPercent 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