What weight of oxygen is required to completely burn 19.8 g of octane?

Answer 1

The chemical equation for complete burning of octane is:

2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O.

This equation shows that 25 moles of diatomic oxygen are required to completely burn each two moles of octane. The gram molecular mass of octane is 114.23 and the gram molecular mass of diatomic oxygen is 2(15.9994). Therefore, the ratio of the mass of oxygen required to completely burn any given mass of octane to the mass of the octane to be burned is 50(15.9994)/2(114.23) or 3.5016, to the justified number of significant digits (the same as the number of digits in the least precisely specified datum 114.23, and the mass of oxygen required to burn 19.8 g of octane is (3.5016)(19.8) or 69.3 grams to the justified number of significant digits, now limited by the less precisely specified datum 119.8.

Answer 2

The balanced chemical equation for the combustion of octane is 2C8H18 + 25O2 --> 16CO2 + 18H2O. This means that 25 moles of O2 are required to completely burn 1 mole of octane. Since the molar mass of O2 is 32 g/mol, the amount of O2 needed to burn 19.8 g of octane is (25 moles O2 / 2 moles C8H18) * (32 g O2 / 1 mole O2) * (19.8 g C8H18).

Therefore, the weight of oxygen required is 400.8 g.