When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3} .\) In a certain experiment, \(20.00 \mathrm{~g}\) iron metal was reacted with \(11.20 \mathrm{~g}\) oxygen gas. After the experiment, the iron was totally consumed, and \(3.24 \mathrm{~g}\) oxygen gas remained. Calculate the amounts of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed in this experiment.

We need to consider the two reactions between iron and oxygen separately. The balanced chemical equations for these reactions are: 1. For the synthesis of FeO: \(2 \: Fe + O_2 \rightarrow 2 \: FeO\) 2. For the synthesis of Fe₂O₃: \(4 \: Fe + 3 \: O_2 \rightarrow 2 \: Fe_2O_3\)

Calculate the moles of iron and oxygen present initially using their molar masses. Here, molar mass of Fe is 55.85 g/mol and molar mass of O₂ is 32.00 g/mol. Moles of Fe = \( \frac{20.00 \: g}{55.85 \: g/mol} = 0.358 \: mol\) Moles of O₂ = \( \frac{11.20 \: g}{32.00 \: g/mol} = 0.350 \: mol\)

Subtract the moles of remaining oxygen gas from moles of initial oxygen to determine the moles of reacted oxygen: Moles of O₂ reacted = Initial moles of O₂ - Remaining moles of O₂ Remaining moles of O₂ = \( \frac{3.24 \: g}{32.00 \: g/mol} = 0.101 \: mol\) Moles of O₂ reacted = 0.350 mol - 0.101 mol = 0.249 mol

Use the stoichiometry of the reactions to determine the moles of FeO and Fe₂O₃ produced. As 0.358 mol of Fe reacted and no Fe was left, the maximum moles of FeO and Fe₂O₃ formed can be: Maximum moles of FeO = \( \frac{0.358 \: mol \: Fe}{2} = 0.179 \: mol\) Maximum moles of Fe₂O₃ = \( \frac{0.358 \: mol \: Fe}{4} = 0.0895 \: mol\) Using these values and the stoichiometry of equation 1, we can find the amount of oxygen used for FeO formation. Moles of O₂ used for FeO formation = \(0.179 \: mol\) Since we have 0.249 mol of O₂ reacted, the remaining oxygen must be consumed for Fe₂O₃ formation. Moles of O₂ used for Fe₂O₃ formation = Moles of O₂ reacted - Moles of O₂ used for FeO formation = 0.249 mol - 0.179 mol = 0.070 mol Now, using the stoichiometry of equation 2, calculate the moles of Fe₂O₃ formed: Moles of Fe₂O₃ formed = \( \frac{0.070 \: mol \: O₂}{3} \times 2 = 0.0467 \: mol\)

Use the molar masses of FeO and Fe₂O₃ to calculate the mass of each compound formed: Mass of FeO formed = Moles of FeO × Molar mass of FeO = 0.179 mol × (55.85 g/mol + 16.00 g/mol) = 0.179 mol × 71.85 g/mol = 12.86 g Mass of Fe₂O₃ formed = Moles of Fe₂O₃ × Molar mass of Fe₂O₃ = 0.0467 mol × (2 × 55.85 g/mol + 3 × 16.00 g/mol) = 0.0467 mol × 159.70 g/mol = 7.46 g

In this experiment, 12.86 g of FeO and 7.46 g of Fe₂O₃ were formed.