Chemistry Q51 – JEE-MAIN 2026

Home Chemistry Basic ChemistryQuestion #34

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 :

Clue 1 of 3: Stoichiometry Basics

Identify the balanced chemical reaction between iron and steam to form $F e_{3} O_{4}$ .

🧠 Full Solution Path

Step 1: Write the balanced chemical equation✦ Active

The reaction between iron and steam to form $F e_{3} O_{4}$ is:

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

Step 2: Calculate moles of steam (

H_{2} O

)○ Expand

Molar mass of $H_{2} O = (2 \times 1) + 16 = 18 {g mol}^{- 1}$ . Moles of $H_{2} O = \frac{Mass of H_{2} O}{Molar mass of H_{2} O} = \frac{18 g}{18 {g mol}^{- 1}} = 1 mol$ .

Step 3: Calculate mass of iron converted○ Expand

From the balanced equation, $4 moles of H_{2} O$ react with $3 moles of F e$ . Therefore, $1 mole of H_{2} O$ reacts with $\frac{3}{4} moles of F e$ . Moles of $F e$ converted $= \frac{3}{4} \times 1 mol = 0.75 mol$ . Molar mass of $F e = 56 {g mol}^{- 1}$ . Mass of $F e$ converted $= Moles of F e \times Molar mass of F e = 0.75 mol \times 56 {g mol}^{- 1} = 42 g$ .

💡 Teacher's Secret Hint

Ensure to use the correct molar masses and stoichiometric ratios from the balanced equation.

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Chemistry Question 51 – JEE-MAIN 2026

Clue 1 of 3: Stoichiometry Basics

🧠 Full Solution Path