12_3 review

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12.3 Limiting Reagent and
Percent Yield
B Y:
MICHELLE ALICE GANIAN
Limiting Reagent
 In a chemical reaction, an insufficient quantity of any of the reactants will
limit the amount of product that is formed.
 (In terms of ingredients, the smallest amount you have of the ingredients
will be the limiting reagent, because it is the substance that will limit the
amount of the recipe.)
 Limiting Reagent-the reagent that determines the amount of product that
can be formed by the reaction. When the limiting reagent is used up, the
chemical reaction is over.
 Excess reagent-the reaction that is not completely used up in an equation.
Finding the Limiting Reagent
 How many grams of metallic copper can be obtained when 54g
Al react with 319g of CuSO4?
 2Al + 3CuSO4 1Al2(SO4)3 + 3Cu
 What is the limiting reagent?
 Step 1: Change grams to moles to find the limiting reagent
 54g Al x 1 mole Al/27g Al = 2 mole Al (you have 2Al in the equation)
 319g CuSO4 x I mole CuSO4/159.5g CuSO4 = 2 mole CuSO4
Using the Limiting Reagent
 How many grams of metallic copper can be obtained when 54g
Al react with 319g of CuSO4?
 Step 2: Continue the problem by using the founded limiting
reagent to solve for the quantity of the product
 Limiting reagent = 2 mole CuSO4
 2 mole CuSO4 x 3 mole Cu/3 mole CuSO4 x 63.5g Cu/1 mole Cu =
127g Cu
Percent Yield
 Actual yield - the amount of product that actually forms
when a reaction is carried out in a lab
 Theoretical yield- the maximum amount of product that
can be formed from given amounts of reactants
 Percent yield – the ratio of the actual yield to the
theoretical yield expressed as a percent.
Calculating Theoretical Yield
 What is the theoretical yield of CaO if 24.8g of CaCO3 is
heated?

CaCO3  CaO3 + CO2
Given mass CaCO3 = 24.8g
Molar mass CaCO3 = 100.1 g
Molar mass CaO = 56.1g
Theoretical yield is unknown.
24.8g CaCO3 x 1 mole CaCO3/100.1g CaCO3 x 1 mole CaO/1 mole
CaCO3 x 56.1g CaO/ 1 mole CaO = 13.9g CaO
Calculating for Percent Yield
 What is the percent yield if 13.1g CaO is actually produced
when 24.8g CaCO3 is heated?
 (You don’t need 24.8 g CaCO3)
 Percent yield = actual yield/theoretical yield x 100%
 Percent yield =13.1g CaO/13.9g CaO x 100% = 94.2
 Hint: the larger number always goes as the denominator

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