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 volume of `H_(2)` liberated at STP (when molar volume of `H_(2) " is " 22.4 L`)

Explanation

To solve the problem step by step, we will follow the stoichiometry of the reaction provided and use the given data to find the volume of hydrogen gas liberated at STP.

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 3 \text{Fe} + 4 \text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4 \text{H}_2(g) \]

### Step 2: Calculate the moles of iron consumed
We know the mass of iron consumed is 56 g. The molar mass of iron (Fe) is approximately 56 g/mol.

To find the moles of iron:
\[ \text{Moles of Fe} = \frac{\text{Mass of Fe}}{\text{Molar mass of Fe}} = \frac{56 \text{ g}}{56 \text{ g/mol}} = 1 \text{ mol} \]

### Step 3: Use stoichiometry to find moles of hydrogen gas produced
From the balanced equation, we see that 3 moles of iron produce 4 moles of hydrogen gas. Therefore, we can set up a ratio:
\[ \frac{4 \text{ moles of } H_2}{3 \text{ moles of Fe}} = \frac{x \text{ moles of } H_2}{1 \text{ mole of Fe}} \]
Solving for \(x\):
\[ x = \frac{4}{3} \text{ moles of } H_2 \]

### Step 4: Calculate the volume of hydrogen gas at STP
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L.

Using the moles of hydrogen gas calculated:
\[ \text{Volume of } H_2 = \text{Moles of } H_2 \times 22.4 \text{ L/mol} \]
Substituting the moles:
\[ \text{Volume of } H_2 = \left(\frac{4}{3} \text{ moles}\right) \times 22.4 \text{ L/mol} = \frac{4 \times 22.4}{3} \text{ L} \]
Calculating this gives:
\[ \text{Volume of } H_2 = \frac{89.6}{3} \text{ L} \approx 29.87 \text{ L} \]

### Final Answer
The volume of hydrogen gas liberated at STP is approximately **29.87 liters**.

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