If $\frac{1}{2}$ moles of oxygen combine with aluminum to form $A l_{2} O_{3}$ , then the weight of aluminum metal used in the reaction is:
$2 A l + \frac{3}{2} O_{2} ⟶ {A l}_{2} O_{3}$

27 gm

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18 gm

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54 gm

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40.5 gm

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Solution

The correct option is B 18 gm
$2 A l + \frac{3}{2} O_{2} \to A l_{2} O_{3}$

$\frac{3}{2} m o l O_{2}$ requires $2 m o l A l$

$1 m o l O_{2}$ requires $\frac{2 \times 2}{3} m o l A l$

$\frac{1}{2} m o l O_{2}$ requires $\frac{4}{3} \times \frac{1}{2} m o l A l$

$= \frac{2}{3} m o l$

$= \frac{2}{3} \times 27 = 18 g m$ aluminum.

Hence, option B is correct.

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