Limiting Reactant problem

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.

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