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To solve this problem, we'll write the chemical equations for the described transformations: 1. **Iron to Ferric Chloride:** \[ 2 \text{Fe} + 3 \text{Cl}_2 \rightarrow 2 \text{FeCl}_3 \] 2. **Ferric Chloride to Ferric Hydroxide:** \[ \text{FeCl}_3 + 3 \text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3 \text{NaCl} \] 3. **Ferric Hydroxide to Ferric Oxide:** \[ 2 \text{Fe(OH)}_3 \rightarrow \text{Fe}_2\text{O}_3 + 3 \text{H}_2\text{O} \] 4. **Ferric Oxide to Iron(III) Sulfate:** \[ \text{Fe}_2\text{O}_3 + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 3 \text{H}_2\text{O} \] Additionally, you are asked to find the mass of hydrogen that is released when 44.8 liters of hydrogen is produced in one of the reactions. The molar volume of an ideal gas is 22.4 L/mol at standard conditions, so the amount of hydrogen is: - Volume of H₂ given: 44.8 L. - Molar volume at STP: 22.4 L/mol. - Moles of H₂ = 44.8 L / 22.4 L/mol = 2 mol. The mass of hydrogen (\(H_2\)) is: - Molar mass of H₂ = 2 g/mol. - Mass of 2 moles of H₂ = 2 mol × 2 g/mol = 4 g. Therefore, the mass of the released hydrogen is 4 grams.