What mass of $C O_{2}$ could be formed by the reaction of $16$ : g $C H_{4}$ with $48$ g of $O_{2}$ ?
$C H_{4} + 2 O_{2} ⟶ C O_{2} + 2 H_{2} O$

44

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33

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16

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24

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Solution

The correct option is C $33$ g
$C H_{4} + 2 O_{2} ⟶ C O_{2} + 2 H_{2} O$
$1 m o l 2 m o l (44 g)$
$16 g 64 g$
$16$ g $48 g = 1.5 m o l$ (Given)
Thus, $O_{2}$ is the limiting reactant.
$∴ 2$ mol $O_{2}$ give $44$ g $C O_{2}$ .
$1.5$ mol $O_{2}$ give $\frac{44}{5} \times 1.5 = 33$ g $C O_{2}$ .

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