stoichiometric calculations stoichiometry – ch. 9
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Stoichiometric Stoichiometric
CalculationsCalculations
Stoichiometric Stoichiometric
CalculationsCalculations
Stoichiometry – Ch. 9
StoichiometryStoichiometryGreek for “measuring elements”The calculations of quantities in
chemical reactions based on a balanced equation.
We can interpret balanced chemical equations several ways.
Stoichiometry DefinitionStoichiometry DefinitionComposition stoichiometry
deals with the mass relationships of elements in compounds.
Reaction stoichiometry involves the mass relationships between reactants and products in a chemical reaction.
Click below to watch the Visual Concept.
Visual Concept
StoichiometryStoichiometry
A. Proportional RelationshipsA. Proportional Relationships
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 RelationshipsRelationshipsMole RatioMole Ratio is a conversion factor
that relates the amounts in moles of any two substances involved in a chemical reaction
Example: 2Al2O3(l) → 4Al(s) + 3O2(g)
Mole Ratios: 2 mol Al2O3 2 mol Al2O3 4 mol Al
4 mol Al 3 mol O2 3 mol O2
B. 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◦Molarity - moles liters soln◦Molar volume - moles liters gas
4. Check answer.
Converting Between Amounts in Converting Between Amounts in MolesMoles
Conversions of Quantities in Conversions of Quantities in MolesMoles
D. Stoichiometry ProblemsD. Stoichiometry ProblemsHow 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
Periodic Table
MolesA MolesB Massg B
Periodic Table
Balanced Equation
Massg A
• Decide where to start based on the units you are given• Stop based on what unit you are asked for
D. Stoichiometry ProblemsD. Stoichiometry ProblemsHow 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
For example...For example...If 10.1 g of Fe are added to a
solution of Copper (II) Sulfate, how much solid copper would form?
2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu
10.1 g Fe
55.85 g Fe
1 mol Fe
2 mol Fe
3 mol Cu
1 mol Cu
63.55 g Cu
= 17.2 g Cu
How do you get good at How do you get good at this?this?
A. Limiting ReactantsA. Limiting ReactantsAvailable 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 ReactantsThe limiting reactant is the reactant
that limits the amount of the other reactant that can combine and the amount of product that can form in a chemical reaction.
The excess reactant is the substance that is not used up completely in a reaction.
How do you find out?How do you find out?Do two stoichiometry problems.
The one that makes the least product is the limiting reagent.
Click below to watch the Visual Concept.
Visual Concept
Limiting Reactants and Limiting Reactants and Excess ReactantsExcess Reactants
If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?
2Cu + S Cu2S
10.6 g Cu 63.55g Cu
1 mol Cu
2 mol Cu
1 mol Cu2S
1 mol Cu2S
159.16 g Cu2S
= 13.3 g Cu2S
3.83 g S 32.06g S
1 mol S
1 mol S
1 mol Cu2S
1 mol Cu2S
159.16 g Cu2S
= 19.0 g Cu2S
= 13.3 g Cu2S
Cu is Limiting Reagent
Percentage Yield Percentage Yield The theoretical yield is the
maximum amount of product that can be produced from a given amount of reactant.
The actual yield of a product is the measured amount of that product obtained from a reaction.
.
Section 3 Limiting Reactants and Percentage Yield
Percent yieldPercent yield = Actual x 100 % Theoretical
Percent YieldPercent YieldWhen 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 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 gTheoretical Yield:
Percent 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