Answer:
a) %yield= 33.00 %
b) %yield= 72.1 %
Explanation:
Step 1: Data given
Mass of iron(III) oxide = 65.0 grams
mass of iron produced = 15.0 grams
Molar mass of Fe2O3 = 159.69 g/mol
Molar mass of CO = 44.01 g/mol
Step 2: The balanced equation:
Fe2O3(s)+3CO(g) →2Fe(s)+3CO2(g)
Step 3: Calculate moles of Fe2O3
Moles Fe2O3 = 65.0 grams / 159.69 g/mol
Moles Fe2O3 = 0.407 moles
Step 4: Calculate moles Fe
For 1 mole Fe2O3 we'll have 2 moles Fe
For 0.407 moles Fe2O3 we'll have 2*0.407 = 0.814 molesFe
Step 5: Calculate mass Fe
Mass fe = 0.814*55.845 g/mol
Mass Fe = 45.46 grams = theoretical yield
Step 6: Calculate % yield
%yield = (actual yield/theoretical yield)*100%
%yield = (15.0/45.46)*100%
%yield= 33.00 %
b) What is the percent yield for the reaction if 75.0 g of carbon monoxide produces 85.0 g of carbon dioxide?
Step 1: Data given
Mass of CO = 75.0 grams
mass of CO2 produced = 85.0 grams
Molar mass of CO = 28.01 g/mol
Molar mass of CO2 = 44.01 g/mol
Step 2: The balanced equation:
Fe2O3(s)+3CO(g) →2Fe(s)+3CO2(g)
Step 3: Calculate moles of CO
Moles CO = 75.0 grams / 28.01 g/mol
Moles CO = 2.68 moles
Step 4: Calculate moles CO2
For 1 mole Fe2O3 we need 3 moles CO to produce 2 moles Fe and 3 moles CO2
For 2.68 moles CO we'll have 2.68 moles CO2
Step 5: Calculate mass CO2
Mass CO2= 2.68 * 44.01 g/mol
Mass CO2 = 117.95 grams = theoretical yield
Step 6: Calculate % yield
%yield = (actual yield/theoretical yield)*100%
%yield = (85.0/117.95)*100%
%yield= 72.1 %
