Calculate the energy change (in kJ) for the combustion of 3.02 grams of ethane using 12.95 grams of oxygen.

Answer 1

To answer this you need the enthalpy of combustion for ethane. I found a value of -1561kJ/mole. Your mileage may vary.

2C₂H₆ + 7O₂ ==> 2CO_{2 +} 3H₂O

3.02 g ethane x 1 mole ethane/30 g = 0.100 moles ethane

12.95 g O₂ x 1 mole/32 g = 0.405 moles O₂

Limiting reactant:

0.1 mole ethane/2 = 0.05

0.405 mole Oxygen/7 = 0.058

Ethane would be limiting

0.1 mole ethane x 1561 kJ/mole = 156 kJ = energy change