The mass of iron converted into by the action of 18 g of steam is

Question

The mass of iron converted into ${Fe}_{3} O_{4}$ by the action of 18 g of steam is:

(Given: Molar mass of H, O and Fe are 1, 16 and 56 g ${mol}^{- 1}$ respectively)

Assume iron is present in excess. [2026]

Smart Achievers · Accepted Answer

(4)

$Moles of steam = \frac{Given mass of H_{2} O}{Molar mass of H_{2} O}$

$= \frac{18 g}{18 g {mol}^{- 1}} = 1 mol$

$3 F e (s) + 4 H_{2} O (g) ⟶ F e_{3} O_{4} (s) + 4 H_{2} (g)$

$By stoichiometry:$

$\frac{n_{F e}}{3} = \frac{n_{H_{2} O}}{4} \Rightarrow \frac{n_{F e}}{3} = \frac{1}{4}$

$n_{F e} = \frac{3}{4}$

$Mass of Fe = Moles of Fe \times Molar mass of Fe$

$= \frac{3}{4} \times 56 g = 42 g$

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