Calculate the mass of iron which will be converted into its oxide (`Fe_3O_4`) by the action of 26g of steam on it. [At mass of Fe=56] `Fe+H_2​O→Fe_3O_4+4H_2` ​`

Options

A. 61g
B. 66g
C. 18g
D. 20g

Explanation

To solve the problem of calculating the mass of iron that will be converted into its oxide (Fe₃O₄) by the action of 26 g of steam, we can follow these steps:

### Step 1: Write the Balanced Chemical Equation
The balanced chemical reaction is:
\[ 3 \text{Fe} + 4 \text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4 \text{H}_2 \]

### Step 2: Calculate the Molar Mass of Reactants
- **Molar mass of Iron (Fe)**:
\[ \text{Molar mass of Fe} = 56 \, \text{g/mol} \]
Therefore, for 3 moles of Fe:
\[ 3 \times 56 = 168 \, \text{g} \]

- **Molar mass of Water (H₂O)**:
\[ \text{Molar mass of H}_2\text{O} = 2 \times 1 + 16 = 18 \, \text{g/mol} \]
Therefore, for 4 moles of H₂O:
\[ 4 \times 18 = 72 \, \text{g} \]

### Step 3: Establish the Stoichiometric Ratio
From the balanced equation, we see that:
- 72 g of steam (H₂O) reacts with 168 g of iron (Fe).

### Step 4: Calculate the Mass of Iron Required for 26 g of Steam
We need to find out how much iron reacts with 26 g of steam. We can set up a proportion based on the stoichiometry:
\[ \frac{168 \, \text{g of Fe}}{72 \, \text{g of H}_2\text{O}} = \frac{x \, \text{g of Fe}}{26 \, \text{g of H}_2\text{O}} \]

Cross-multiplying gives:
\[ x = \frac{168 \times 26}{72} \]

### Step 5: Calculate the Value of x
Now, calculate the value:
\[ x = \frac{168 \times 26}{72} = \frac{4368}{72} \approx 60.67 \, \text{g} \]

### Step 6: Round to Appropriate Significant Figures
Rounding 60.67 g gives approximately:
\[ x \approx 61 \, \text{g} \]

### Final Answer
The mass of iron that will be converted into its oxide by the action of 26 g of steam is approximately **61 g**.

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