Answer:
Option A (233.3 gram)
Explanation:
Step 1: Data given
Mass of C3H6 produced = 91.3 %
% yield = 81.3 %
Step 2: The balanced equation
2C3H6+ 9O2 → 6CO2+ 6H2O
Step 3: Calculate moles C3H6
Moles C3H6 = mass C3H6 / molar mass C3H6
Moles C3H6 = 91.3 grams / 42.08 g/mol
Moles C3H6 = 2.17 moles
Step 4: Calculate moles CO2
For 2 moles C3H6 we need 9 moles O2 to produce 6 moles CO2 and 6 moles H2O
For 2.17 moles C3H6 we'll have 3*2.17 = 6.51 moles CO2
Step 5: Calculate mass CO2
Mass CO2 = moles CO2 * molar mass CO2
Mass CO2 = 6.51 moles * 44.01 g/mol
Mass CO2 = 286.5 grams
Step 6: Calculate actual yield of moles CO2
% yield = (actual yield / theoretical yield ) *100 %
actual yield = % yield * theoretical yield / 100 %
actual yield = (81.3 % * 6.51 moles ) / 100 %
actual yield = 5.29 moles
Step 7: Calculate actual yield of mass CO2
Mass CO2 = moles * molar mass
Mass CO2 = 5.29 * 44.01 g/mol
Mass CO2 = 233 grams
Option A (233.3 gram) is the clostest
