Answer:
15.6 liters of oxygen gas at STP are required to completely react with 52.0 g of iron to form iron(III) oxide
Explanation:
Let’s calculate the volume of oxygen gas required to completely react with 52.0 g of iron (Fe) to form iron(III) oxide (Fe<sub>2</sub>O<sub>3</sub>).
First, we need to find the moles of iron (Fe):
The molar mass of iron (Fe) is approximately 55.85 g/mol.
Moles of Fe = Mass of Fe / Molar mass of Fe Moles of Fe = 52.0 g / 55.85 g/mol Moles of Fe ≈ 0.931 mol
Next, let’s determine the stoichiometric ratio between iron (Fe) and oxygen (O<sub>2</sub>) in the balanced chemical equation:
The balanced equation for the reaction is: [ 4Fe + 3O_2 \rightarrow 2Fe_2O_3 ]
From the equation, we see that 4 moles of Fe react with 3 moles of O<sub>2</sub>.
Therefore, the moles of O<sub>2</sub> required = (3/4) × Moles of Fe Moles of O<sub>2</sub> required ≈ (3/4) × 0.931 mol Moles of O<sub>2</sub> required ≈ 0.698 mol
Now, let’s find the volume of oxygen gas at STP (Standard Temperature and Pressure):
At STP, 1 mole of any gas occupies a volume of 22.4 L.
Volume of O<sub>2</sub> required = Moles of O<sub>2</sub> × Molar volume at STP Volume of O<sub>2</sub> required ≈ 0.698 mol × 22.4 L/mol Volume of O<sub>2</sub> required ≈ 15.6 L
Therefore, approximately 15.6 liters of oxygen gas at STP are required to completely react with 52.0 g of iron to form iron(III) oxide.