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].
