Question

Balance the following chemical equation, then answer the following question.

C8H18(g) + O2(g) →CO2(g)+H2O(g)
How many grams of oxygen are required to react with 10.0 grams of octane (C8H18) in the combustion of octane in gasoline?

How do I do this?

Answer 1

Molar mass:

O2 = 31.99 g/mol
C8H18 = 144.22 g/mol

2 C8H18(g) + 25 O2(g) = 16 CO2(g) + 18 H2O(g)

2 x 144.22 g --------------- 25 x 31.99 g
10.0 g ----------------------?? ( mass of O2)

10.0 x 25 x 31.99 / 2 x 144.22 =

7997.5 / 288.44 => 27.72 g of O2

hope this helps!