30 grams of pentane (c5h12) undergo combustion. How many grams of oxygen were consumed

Answer 1

Assuming complete combustion of the pentane, the balanced chemical equation will be:

C5H12 + 8O2 → 5CO2 + 6H2O

You will need a balanced chemical equation to get the mole ratio between the pentane that you have (30g) and the O2 that you want.

Conversion factors required:

Molar masses of:

Pentane: 72.17 g C5H12 = 1 mole C5H5

Oxygen: 32.00 g O2 = 1 mole O2

Mole ratio of pentane to oxygen:

1 mole C5H12 = 8 moles O2

Solution:

30 g C5H12 * 1 mole C5H12/72.17g C5H12 * 8 moles O2/1 mole C5H12 * 32.00 g O2/1 mole O2= 106 g O2