Assuming that petrol is octane $\left({\mathrm{C}}_8{\mathrm{H}}_{18}\right)$ and has density 0.8 g/ml, 1.425 litre of petrol on complete combustion will consume

$\begin{array}{l}\mathrm{Mass}\mathrm{of}\mathrm{octane}\text{\hspace{0.17em}=d×volume\hspace{0.17em}}=1.425\times 103\times 0.8\mathrm{g}\\ \mathrm{Moles}\mathrm{of}\mathrm{octane}=\frac{1425\times 8}{114}=10\mathrm{moles}\\ {\mathrm{C}}_8{\mathrm{H}}_{18}+\frac{25}{2}{\mathrm{O}}_2\stackrel{ }{\to }8{\mathrm{CO}}_2+9{\mathrm{H}}_2\mathrm{O}\end{array}$

From the equation it can be seen

For 1 mole octane required $\frac{25}{2}\mathrm{moles}$ of oxygen.

$\therefore \mathrm{for}10\mathrm{moles}\mathrm{octane}\mathrm{required}\frac{25}{2}\times 10=125\mathrm{moles} \mathrm{of} \mathrm{oxygen}$