JEE Main 2026 — Apr 2, Shift 1 — Chemistry solution

JEE Main2026Apr 2, Shift 1Chemistry

The mass of iron converted into $F e_{3} O_{4}$ 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 :

Solution

The balanced reaction is:

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

Moles of steam $= 18/18 = 1$ mol.

From stoichiometry, $4$ mol steam needs $3$ mol Fe, so $1$ mol steam needs $3/4$ mol Fe.

Mass of Fe required $= \frac{3}{4} \times 56 = 42$ g.

Practice more Chemistry

Concept-wise practice with instant solutions on Fundamenthol.