Calculate the weight of Iron which will be converted into its oxide by the action of 18 g of steam.
Given the reaction :
$3 F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}$
(Atomic mass of Fe=56 g.

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Solution

The balanced equation is
$3 F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}$
Moles of steam = $\frac{m a s s}{M o l a r m a s s}$
$\frac{18}{18} = 1 m o l e$
Moles of Fe= $\frac{3}{4} \times m o l e s o f H_{2} O$
= $\frac{3}{4} \times 1.0 = 0.75$
Mass of Fe = $m o l e \times m o l a r m a s s$
$0.75 \times 56 = 42$
= $42 g$

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Similar questions

(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 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

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