A quantity of 8 g $C H_{4}$ is mixed with 28 g $O_{2}$ and fired. Which of the following are correct about the combustion of $C H_{4}$ in this condition?

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1 g $C H_{4}$ will remain left unburned if carbon is quantitatively converted into $C O_{2}$ .

18 g water will form in any possible condition.

4 g $O_{2}$ will remain unused if carbon is quantitatively converted into CO.

Equal moles of CO and $C O_{2}$ are formed if none of the reactants is left and there is no other side reaction.

answer is A, B, C.

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Detailed Solution

A) $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$
$16 g 64 g$
$7 g$ 28 g
B) $C H_{4} + \frac{3}{2} O_{2} \to C O + 2 H_{2} O$
$16 g 48 g$
$∴ 8 g 24 g$

$\frac{1}{2} C H_{4} + \frac{7}{8} O_{2} \to \frac{1}{4} C O_{2} + \frac{1}{4} C O + H_{2} O$

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