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The weight of iron which will be converted into its oxide ( ${Fe}_{3} O_{4}$ ) by the action of 18 g of steam on it will be (Atomic weight of Fe = 56)

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168 g

84 g

42 g

21 g

answer is C.

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Detailed Solution

$3 Fe + 4 H_{2} O \to {Fe}_{3} O_{4} + 4 H_{2}$

3 $\times$ 56g 4 $\times 18 g$

= 168 g = 72 g

$∵ 72 g steam required Fe = 168 g$

$∴ 18 g steam required Fe = \frac{168 \times 18}{72} = 42 g$

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