If $1 \frac{1}{2}$ moles of oxygen combine with Al to form ${Al}_{2} O_{3}$ the weight of Al used in the reaction is (Al = 27)

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54 g

27 g

49.5 g

31 g

answer is B.

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Detailed Solution

$\begin{array}{l} 2 Al + \frac{3}{2} O_{2} \to {Al}_{2} O_{3} \\ According to equation \frac{3}{2} mole of O_{2} combines with 2 moles Al . \\ 2 mole Al = 54 g \end{array}$

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