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2 moles of ${\text{O}}_{2}\text{+}7.52$ moles of N2 = 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: $A:F=\overline{A:F}\left(\frac{{M}_{\text{air}}}{{M}_{\text{fuel}}}\right)$ and M is the molecular weight.

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

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