The mass of Fe converted to its oxide $F e_{3} O_{4}$ by the action of 18 g of steam is:

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Solution

$3 F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}$
$3 \times 56$ g of Fe requires $4 \times 18$ g of $H_{2} O$
Mass of Fe requires for 18 g of $H_{2} O$ is = $\frac{(3 \times 56 \times 18)}{(4 \times 18)} = 42 g$

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Similar questions

The mass of Fe converted to its oxide $F e_{3} O_{4}$ by the action of 18 g of steam is:

3 F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}

F e = 56 g

(F e_{3} O_{4})

8 g

F e = 56

F e + H_{2} O \to F e_{3} O_{4} + 4 H_{2}

18

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