Answer:
Since ethane is the limiting reactant. It will completely be consumed. There will remain 0 grams of ethane.
Explanation:
Step 1: Data given
Mass of C2H6 = 20.1 grams
Mass of O2 = 95.0 grams
Molar mass ethane = 30.07 g/mol
Molar mass O2 = 32.0 g/mol
Molar mass CO2 = 44.01 g/mol
Step 2: The balanced equation
2C2H6 + 7O2 → 4CO2 + 6H2O
Step 3: Calculate moles C2H6
Moles C2H6 = mass C2H6 / molar mass C2H6
Moles C2H6 = 20.1 grams /30.07 g/mol
Moles C2H6 = 0.668 moles
Step 4: Calculate moles O2
Moles O2 = 95.0 grams / 32.0 g/mol
Moles O2 = 2.97 moles
Step 5: Calculate limiting reactant
For 2 moles C2H6 we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O
C2H6 is the limiting reactant. It will completely be consumed ( 0.668 moles).
O2 is in excess. There will react 7/2 * 0.668 = 2.338 moles
There will remain 2.97 - 2.338 = 0.632 moles O2
Since ethane is the limiting reactant. It will completely be consumed. There will remain 0 grams of ethane.
