W is the mass of iron (in g) which will be converted into `Fe_(3)O_(4)` by the action of 18 g of steam on it . What is the value of W/7 ?
`Fe+H_(2)O to Fe_(3)O_(4)+H_(2)`

Answer

To solve the problem, we need to find the mass of iron (W) that will be converted into `Fe_(3)O_(4)` by the action of 18 g of steam (water). The balanced chemical equation for the reaction is:

\[ 3 \text{Fe} + 4 \text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4 \text{H}_2 \]

### Step-by-step Solution:

1. **Identify the Molar Masses:**
- Molar mass of Fe (Iron) = 56 g/mol
- Molar mass of H2O (Water) = 18 g/mol
- Molar mass of Fe3O4 = (3 × 56) + (4 × 16) = 168 + 64 = 232 g/mol

2. **Determine Moles of Water:**
- Given mass of steam (water) = 18 g
- Moles of H2O = \(\frac{18 \text{ g}}{18 \text{ g/mol}} = 1 \text{ mol}\)

3. **Use Stoichiometry to Find Moles of Iron Required:**
- According to the balanced equation, 4 moles of H2O react with 3 moles of Fe.
- Therefore, 1 mole of H2O will react with \(\frac{3}{4}\) moles of Fe.
- Moles of Fe required = \(1 \text{ mol H2O} \times \frac{3 \text{ mol Fe}}{4 \text{ mol H2O}} = \frac{3}{4} \text{ mol Fe}\)

4. **Convert Moles of Iron to Mass:**
- Mass of Fe (W) = Moles of Fe × Molar mass of Fe
- \(W = \frac{3}{4} \text{ mol} \times 56 \text{ g/mol} = 42 \text{ g}\)

5. **Calculate W/7:**
- \(W/7 = \frac{42 \text{ g}}{7} = 6 \text{ g}\)

### Final Answer:
The value of \(W/7\) is **6 g**.