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22. Iron reacts with steam according to the equation:
[tex]\[ 3 \text{Fe} (s) + 4 \text{H}_2\text{O} (g) \rightarrow \text{Fe}_3\text{O}_4 (s) + 4 \text{H}_2 (g) \][/tex]

Calculate the mass of iron required to produce 2.24 L of hydrogen at STP. (3 marks)


Sagot :

To determine the mass of iron required to produce 2.241 liters of hydrogen (H₂) at Standard Temperature and Pressure (STP), we need to follow several steps involving the concepts of stoichiometry and gas laws.

### Step-by-Step Solution:

Step 1: Calculate Moles of Hydrogen Gas (H₂) Produced

At STP (Standard Temperature and Pressure), 1 mole of any ideal gas occupies 22.414 liters. Given that we have 2.241 liters of H₂, we can calculate the number of moles of H₂ using the molar volume.

[tex]\[ \text{Moles of } H_2 = \frac{\text{Volume of } H_2}{\text{Molar Volume of gas at STP}} = \frac{2.241 \text{ L}}{22.414 \text{ L/mol}} \approx 0.09998 \text{ mol} \][/tex]

Step 2: Use Stoichiometry to Determine Moles of Iron (Fe) Required

The balanced chemical equation for the reaction is:
[tex]\[ 3 \text{ Fe } + 4 \text{ H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4 \text{ H}_2 \][/tex]

From the balanced equation, we see that 3 moles of Fe produce 4 moles of H₂. Thus, we can set up the stoichiometric ratio to find the moles of Fe needed:

[tex]\[ \text{Moles of } Fe = \left( \frac{3 \text{ moles of Fe}}{4 \text{ moles of } H_2} \right) \times \text{Moles of } H_2 = \left( \frac{3}{4} \right) \times 0.09998 \approx 0.07499 \text{ mol} \][/tex]

Step 3: Calculate the Mass of Iron Required

The final step is to convert the moles of Fe to grams. The molar mass of iron (Fe) is 55.845 g/mol. Therefore, the mass of Fe needed is:

[tex]\[ \text{Mass of } Fe = \text{Moles of } Fe \times \text{Molar mass of } Fe = 0.07499 \text{ mol} \times 55.845 \text{ g/mol} \approx 4.188 \text{ g} \][/tex]

### Conclusion:

The mass of iron required to produce 2.241 liters of hydrogen gas at STP is approximately [tex]\(4.188 \text{ grams}\)[/tex].
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