Answer:
The percent yield = 61.8 %
Explanation:
Step 1: Data given
Mass of CO2 produced = 12.4 grams
Mass of methane = 7.38 grams
Mass of oxygen = 39.2 grams
Molar mass of CO2 = 44.01 g/mol
Molar mass of methane = 16.04 g/mol
Molar mass of oxygen = 32.0 g/mol
Step 2: The balanced equation
CH4 + 2O2 → CO2 + 2H2O
Step 3: Calculate moles methane
Moles methane = mass methane / molar mass methane
Moles methane = 7.38 grams / 16.04 g/mol
Moles methane = 0.460 moles
Step 4: Calculate moles oxygen
Moles O2 = 39.2 grams / 32.0 g/mol
Moles O2 = 1.225 moles
Step 5: Calculate limiting reactant
For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O
Methane is the limiting reactant. It will completely be consumed (0.460 moles). O2 is in excess. There will react 2*0.460 = 0.920 moles
There will remain 1.225 -0 .920 = 0.305 moles O2
Step 6: Calculate moles of CO2
For 1 mol CH4 we need 2 moles O2 to produce 1 mol CO2 and 2 moles H2O
For 0.460 moles CH4 we'll have 0.460 moles CO2
Step 7: Calculate mass CO2
Mass CO2 = moles CO2 * molar mass CO2
Mass CO2 = 0.460 moles * 44.01 g/mol
Mass CO2 = 20.24 grams
Step 8: Calculate percent yield
% yield = (actual yield/theoretical yield) *100%
% yield = (12.5 grams / 20.24 grams ) *100%
% yield = 61.8 %
The percent yield = 61.8 %
