2 moles of ${\text{O}}_{2}\text{+}7.52$ moles of N_{2} = 9.52 moles of air per mole of fuel. Thus,

the air–fuel ratio on a molar basis is $\overline{A:F}=9.52$, and

the air–fuel ratio on a mass basis is $A:F=9.52\left(\frac{28.97}{16.04}\right)=17.19$

where, the relation between *A*:*F* and $\overline{A:F}$ is: $A:F=\overline{A:F}\left({\scriptscriptstyle \frac{{M}_{\text{air}}}{{M}_{\text{fuel}}}}\right)$ and *M* is the molecular weight.

The above equations indicate that the amount of air required ...

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