Answer:
The percent yield is 28.0 %
Explanation:
Step 1: Data given
Mass of carbon dioxide = 2.71 grams
Mass of ethane = 3.31 grams
Mass of oxygen = 23.9 grams
Molar mass ethane = 30.07 g/mol
Molar mass CO2 = 44.01 g/mol
Molar mass O2 = 32.0 g/mol
Step 2: The balanced equation
2C2H6 + 7O2 → 4CO2 + 6H2O
Step 3: Calculate moles
Moles = mass /molar mass
Moles ethane = 3.31 grams / 30.07 g/mol
Moles ethane = 0.110 moles
Moles oxygen = 23.9 grams / 32.0 g/mol
Moles oxygen = 0.747 moles
Step 4: Calculate the limiting reactant
For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O
Ethane is the limiting reactant. It will completely be consumed (0.110 moles)
Oxygen is the limiting reactant. There will react 0.110 *3.5 = 0.385 moles
There will remain 0.747 - 0.385 = 0.362 moles O2
Step 5: Calculate moles CO2
For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O
For 0.110 moles ethane we'll have 2*0.110 = 0.220 moles CO2
Step 6: Calculate theoretical yield of CO2
Mass CO2 = moles CO2 * molar mass CO2
Mass CO2 = 0.220 moles * 44.01 g/mol
Mass CO2 = 9.68 grams
Step 7: Calculate % yield
% yield = (actual yield / theoretical yield) * 100 %
% yield =(2.71 grams / 9.68 grams ) * 100 %
% yield = 28.0 %
The percent yield is 28.0 %
