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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?

Verified Answer :

Willchemistry

Molar mass

C8H18 => 114.23 g/mol
O2 => 31.99 g/mol

2 C8H18 + 25 O2 ---> 16 CO2 + 18 H2O

2 x 114.23 g ---------------- 25 x 31.99 g
10.0 g ----------------------- ( mass of O2)

mass of O2 = 10.0 x 25 x 31.99 / 2 x 114.23

mass of O2 = 7997.5 / 228.46

= 35.00 g of O2

hope this helps!

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