Final answer:
The volume of iron that would be oxidized by 89.6L of oxygen to produce Fe2O3 at STP is 37.8 cm³. This is calculated using the stoichiometry of the reaction 4Fe(s) + 3O₂(g) → 2Fe2O3(s), taking into account the molar volume of gases at STP and the molar mass of iron.
Step-by-step explanation:
To determine the volume of iron that would be completely oxidized by 89.6L of oxygen to produce Fe2O3 (iron(III) oxide) at Standard Temperature and Pressure (STP), we need to use the stoichiometry of the balanced chemical equation for the reaction of iron with oxygen. The balanced chemical equation is:
4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)
This equation shows that 4 moles of iron react with 3 moles of oxygen to form 2 moles of Fe₂O₃. At STP, 1 mole of any gas occupies 22.4 liters, so 89.6L of oxygen is equivalent to 4 moles of oxygen (because 89.6 / 22.4 = 4). According to the stoichiometry of the reaction, 3 moles of oxygen react with 4 moles of iron, so 4 moles of oxygen would completely react with:
(4 moles O₂ / 3 moles O₂) × 4 moles Fe = 5.33 moles of Fe
Given that the molar mass of iron is 55.85 grams per mole, the mass of iron that would be oxidized can be calculated as:
(5.33 moles of Fe) × (55.85 g/mol) = 297.6 grams of Fe
To convert this mass to volume, we would need the density of iron (7.87 g/cm³). The volume of iron can then be found using the formula:
Volume = Mass / Density
Volume of Fe = 297.6 g / 7.87 g/cm³ = 37.8 cm³ of Fe (rounded to three significant figures)
Therefore, 89.6L of oxygen at STP would completely oxidize 37.8 cm³ of iron metal to produce Fe2O3.