# Balancing chemical equations - 1 | Chemical Reactivity

The unbalanced chemical equation of the combustion of octane is:

C8H18 (l) + O2 (g) → CO2 (g) + H2O (l)

Balance the equation.

What mass of O2 will be needed to burn 50.8 g of octane?

How many moles of water are produced?

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

According to the stoechiometry of the balanced equation:
$\frac{{\mathrm{n}}_{\mathrm{C}8\mathrm{H}18}}{2}$ = $\frac{{\mathrm{n}}_{\mathrm{O}2}}{25}$  ⇒ $\frac{{\mathrm{m}}_{\mathrm{O}2}}{{\mathrm{M}}_{\mathrm{O}2}}$ = $\frac{25}{2}$ x $\frac{{\mathrm{m}}_{\mathrm{C}8\mathrm{H}18}}{{\mathrm{M}}_{\mathrm{C}8\mathrm{H}18}}$

mO2 = 178 g

$\frac{{\mathrm{n}}_{\mathrm{C}8\mathrm{H}18}}{2}$ = $\frac{{\mathrm{n}}_{\mathrm{H}2\mathrm{O}}}{18}$ ⇒  nH2O $\frac{18}{2}$ x $\frac{{\mathrm{m}}_{\mathrm{C}8\mathrm{H}18}}{{\mathrm{M}}_{\mathrm{C}8\mathrm{H}18}}$
nH2O = 4.01 mol