The weight of oxygen required to completely react with $27$ g of $A l$ is:

8

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16

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32

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24

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Solution

The correct option is D $24$ gm
The balanced chemical equation is as follows:

$4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$ .
Here, $4$ moles $(4 \times 27 g)$ of aluminum reacts with $3$ moles $(3 \times 32 g)$ of oxygen.
Thus, the weight of oxygen required to completely react with $27$ g of aluminum is $(3 \times 32 g) \times \frac{(27 g)}{(4 \times 27 g)} = 24$ gm

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