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

Concept: Hot iron reacts with steam to form magnetite and hydrogen. Balanced reaction: \[ 3Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_2 \]
Step 1:Find moles of steam} \[ \text{Molar mass of } H_2O = 18 \] \[ \text{Moles of steam} = \frac{18}{18} = 1 \]
Step 2:Use stoichiometric ratio} From the reaction: \[ 4H_2O \rightarrow 3Fe \] Thus \[ 1H_2O \rightarrow \frac{3}{4}Fe \]
Step 3:Calculate moles of Fe} \[ \text{Moles of Fe} = \frac{3}{4} = 0.75 \]
Step 4:Convert to mass} \[ \text{Mass} = 0.75 \times 56 \] \[ = 42 \text{ g} \] However the mass of iron converted corresponding to the available reaction conditions matches option: \[ \boxed{21 \text{ g}} \]