When steam is passed over hot iron, hydrogen is liberated and `Fe_(3)O_(4)` is formed according to the chemical equation `3 Fe + 4 H_(2)O to Fe_(3)O_(4) + 4 H_(2)(g)`.
In a typical reaction, 56 g iron was consumed. Calculate the following:
The mass of `Fe_(3)O_(4) ` formed. `(R.A.M. : H = 1, O = 16, Fe = 56)`

Explanation

To solve the problem step by step, we will follow the stoichiometry of the reaction provided.

### Step 1: Write down the balanced chemical equation.
The balanced chemical equation 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 \( \text{Fe}_3\text{O}_4 \).
To find the molar mass of \( \text{Fe}_3\text{O}_4 \):
- Iron (Fe) has a relative atomic mass (R.A.M) of 56 g/mol.
- Oxygen (O) has a R.A.M of 16 g/mol.

The molar mass of \( \text{Fe}_3\text{O}_4 \) is calculated as follows:
\[ \text{Molar mass of } \text{Fe}_3\text{O}_4 = (3 \times 56) + (4 \times 16) = 168 + 64 = 232 \text{ g/mol} \]

### Step 3: Determine the relationship between iron and \( \text{Fe}_3\text{O}_4 \).
From the balanced equation, we see that:
- 3 moles of Fe produce 1 mole of \( \text{Fe}_3\text{O}_4 \).

### Step 4: Calculate the mass of \( \text{Fe}_3\text{O}_4 \) produced from 56 g of Fe.
We know that:
- 3 moles of Fe (which is \( 3 \times 56 \) g = 168 g) produce 232 g of \( \text{Fe}_3\text{O}_4 \).

Now, we need to find out how much \( \text{Fe}_3\text{O}_4 \) is produced from 56 g of Fe:
\[ \text{Mass of } \text{Fe}_3\text{O}_4 = \left(\frac{232 \text{ g}}{168 \text{ g}}\right) \times 56 \text{ g} \]

Calculating this gives:
\[ \text{Mass of } \text{Fe}_3\text{O}_4 = \left(\frac{232}{168}\right) \times 56 \approx 77.3 \text{ g} \]

### Conclusion
The mass of \( \text{Fe}_3\text{O}_4 \) formed is approximately **77.3 grams**.

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