Calculate the mass of iron which will be converted into its oxide $(F e_{3} O_{4})$ by the action of $8 g$ of steam on it. [At mass of $F e = 56$ ]
$F e + H_{2} O \to F e_{3} O_{4} + 4 H_{2}$

21

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12

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42

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84

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Solution

The correct option is A $21$ g
$3 F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}$
As per stiochiometry, Moles of $H_{2} O$ =Moles of fe
$4 \times \frac{8}{18} = \frac{m}{56} \times 3$
$m = 21 g$
Therefore, mass of iron converted to $F e_{3} O_{4}$ is $21 g .$

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Similar questions

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

F e = 56 g

18 g

18 g

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

56 g/mol

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

18 g

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

56 g/mol

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