how many grams of iron(iii) oxide (Fe2O3) can be produced from 25.0g of iron and an excess of oxygen?
4Fe+O2–>2Fe2O3
To solve this problem, we need to determine the molar ratio between iron (Fe) and iron(III) oxide (Fe2O3) based on the balanced equation:
4Fe + O2 -> 2Fe2O3
The equation tells us that 4 moles of Fe react to produce 2 moles of Fe2O3.
First, we calculate the number of moles of iron (Fe) using its molar mass:
Molar mass of Fe = 55.85 g/mol
25.0 g of Fe = (25.0 g) / (55.85 g/mol) ≈ 0.448 mol
From the balanced equation, we know that 4 moles of Fe react to produce 2 moles of Fe2O3. Therefore, the number of moles of Fe2O3 produced will be half the number of moles of Fe:
Number of moles of Fe2O3 = (0.448 mol) / 4 * 2 ≈ 0.112 mol
Finally, we calculate the mass of Fe2O3 using its molar mass:
Molar mass of Fe2O3 = 159.69 g/mol
Mass of Fe2O3 = (0.112 mol) * (159.69 g/mol) ≈ 17.92 g
Therefore, approximately 17.92 grams of iron(III) oxide (Fe2O3) can be produced from 25.0 grams of iron.