Answer:
There will remain 8.06 grams of ethane
Explanation:
Step 1: Data given
Mass of ethane = 9.32 grams
Mass of oxygen = 12.0 grams
Molar mass ethane = 30.07 g/mol
Molar mass oxygen = 32.00 g/mol
Step 2: The balanced equation
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
Step 3: Calculate moles ethane
Moles ethane = mass ethane / molar mass ethane
Moles ethane = 9.32 grams / 30.07 g/mol
Moles ethane = 0.3099 moles
Step 4: Calculate moles oxygen
Moles oxygen = 12.0 grams / 32.0 g/mol
Moles oxygen = 0.375 moles
Step 5: Calculate the limiting reactant
For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O
O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)
Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles
There will remain 0.375 - 0.107 = 0.268 moles
Step 6: Calculate mass ethane
Mass ethane = moles ethane * molar mass ethane
Mass ethane = 0.268 moles * 30.07 g/mol
Mass ethane = 8.06 grams
There will remain 8.06 grams of ethane
