Calculate the mass of iron which will be converted into its oxide (`Fe_3O_4`) by the action of 28g of steam on it. [At mass of Fe=56] `Fe+H_2​O→Fe_3O_4+4H_2` ​

Explanation

To calculate the mass of iron that will be converted into its oxide \( Fe_3O_4 \) by the action of 28 g of steam, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 3Fe + 4H_2O \rightarrow Fe_3O_4 + 4H_2 \]

### Step 2: Determine the molar masses
- The molar mass of iron (Fe) is given as 56 g/mol.
- The molar mass of water (H₂O) is calculated as:
\[ H_2O = 2 \times 1 + 16 = 18 \text{ g/mol} \]

### Step 3: Calculate the mass of iron and steam in the balanced equation
From the balanced equation, we see that:
- 72 g of water (4 moles of H₂O) reacts with 168 g of iron (3 moles of Fe).

### Step 4: Set up the proportion for the reaction
We need to find out how much iron reacts with 28 g of steam. We can set up a proportion based on the stoichiometry of the reaction:
\[ \frac{168 \text{ g Fe}}{72 \text{ g H}_2O} = \frac{x \text{ g Fe}}{28 \text{ g H}_2O} \]

### Step 5: Solve for \( x \)
Cross-multiplying to solve for \( x \):
\[ x = \frac{168 \text{ g Fe} \times 28 \text{ g H}_2O}{72 \text{ g H}_2O} \]

Calculating \( x \):
\[ x = \frac{168 \times 28}{72} \]
\[ x = \frac{4704}{72} = 65.33 \text{ g Fe} \]

### Conclusion
The mass of iron that will be converted into its oxide \( Fe_3O_4 \) by the action of 28 g of steam is **65.33 g**.

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