Question

The mass of iron converted into \(Fe_3O_4\) 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.

• A. \(2.1\) g
• B. \(4.2\) g
• C. \(21\) g
• D. \(42\) g

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The reaction between iron and steam at high temperature produces magnetic oxide of iron (\(Fe_3O_4\)) and hydrogen gas. We use stoichiometry to relate the mass of steam consumed to the mass of iron reacted.
Step 2: Key Formula or Approach:
1. Balanced Chemical Equation: \(3Fe(s) + 4H_2O(g) \rightarrow Fe_3O_4(s) + 4H_2(g)\)
2. Mole concept: \(n = \frac{\text{Mass}}{\text{Molar Mass}}\)
Step 3: Detailed Explanation:
From the balanced equation:
4 moles of \(H_2O\) react with 3 moles of \(Fe\).
Molar mass of \(H_2O = 2(1) + 16 = 18 \text{ g/mol}\).
Molar mass of \(Fe = 56 \text{ g/mol}\).
Moles of steam (\(H_2O\)) provided = \(\frac{18 \text{ g}}{18 \text{ g/mol}} = 1 \text{ mole}\).
According to the stoichiometry:
4 moles of \(H_2O\) \(\rightarrow\) 3 moles of \(Fe\)
1 mole of \(H_2O\) \(\rightarrow \frac{3}{4}\) mole of \(Fe\)
Mass of iron converted = \((\text{Moles of } Fe) \times (\text{Molar mass of } Fe)\)
\[ \text{Mass} = \frac{3}{4} \times 56 = 3 \times 14 = 42 \text{ g} \]
Step 4: Final Answer:
The mass of iron converted is 42 g.