Calculate the mass of iron which will be converted into its oxide (`Fe_3O_4`) by the action of 33g of steam on it. [At mass of Fe=56] `Fe+H_2​O→Fe_3O_4+4H_2` ​

Answer

To solve the problem of calculating the mass of iron that will be converted into its oxide (Fe₃O₄) by the action of 33g of steam, we can follow these steps:

### Step 1: Write the balanced chemical equation
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 2: Calculate the molar mass of Fe₃O₄ and H₂O
- The molar mass of iron (Fe) is given as 56 g/mol.
- The molar mass of water (H₂O) is calculated as follows:
- H: 1 g/mol × 2 = 2 g/mol
- O: 16 g/mol
- Total for H₂O = 2 + 16 = 18 g/mol

### Step 3: Determine the total mass of reactants
From the balanced equation, we can see that:
- 3 moles of Fe react with 4 moles of H₂O.
- Therefore, the mass of iron that reacts with 72 g of water (4 moles of H₂O) can be calculated:
- Mass of iron = 3 moles × 56 g/mol = 168 g

### Step 4: Set up a proportion to find the mass of iron for 33 g of steam
Since we know that 72 g of steam reacts with 168 g of iron, we can set up a proportion to find the mass of iron (x) that will react with 33 g of steam:
\[ \frac{168 \text{ g Fe}}{72 \text{ g H}_2\text{O}} = \frac{x \text{ g Fe}}{33 \text{ g H}_2\text{O}} \]

### Step 5: Solve for x
Cross-multiplying gives:
\[ 168 \times 33 = 72 \times x \]
\[ 5544 = 72x \]
Now, divide both sides by 72 to solve for x:
\[ x = \frac{5544}{72} = 77 \text{ g} \]

### Conclusion
The mass of iron that will be converted into its oxide by the action of 33 g of steam is **77 g**.

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