Oktana dalam bensin terbakar menurut reaksi: 2C8H18(l)+25O2(g) →16CO2(g)+18H2O(g). Jika 28,5 g oktana dibakar dan menghasilkan gas CO, berapa persen CO2 yang di

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

m C8H18 = 28,5 gr
Mr C8H18 = 8 Ar C + 18 Ar H
= 8(12) + 18(1)
= 96 + 18
= 114

n C8H18 = m / Mr
= 28,5 gr / 114
= 0,25 mol

n CO2 = ( koef. CO2 / koef. C8H18 ) × n C8H18
= 16/2 × 0,25 mol
= 2 mol

Mr CO2 = Ar C + 2 Ar O
= 12 + 2(16)
= 44

m CO2 = Mr × n
= 44 × 2 mol
= 88 gr

persentase dari C8H18
= ( m CO2 / m C8H18 ) × 100%
= ( 88 gr / 28,5 gr ) × 100%
= 308%