Final answer:
To find the moles of Fe2O3 produced from 27.0 grams of iron, first divide the mass of iron by its molar mass to get moles of iron, then use the balanced chemical reaction to find the stoichiometric mole ratio, leading to the calculation of approximately 0.242 moles of Fe2O3.
Step-by-step explanation:
The question asks how many moles of Fe2O3 will be produced from 27.0 grams of iron (Fe) assuming that oxygen (O2) is in excess. To answer this question, let's first determine the molar mass of iron, which is 55.85 grams per mole. Knowing the molar mass, we can convert the given mass of iron to moles by dividing the mass by the molar mass.
27.0 g Fe ÷ 55.85 g/mol Fe = 0.483 mol Fe
Next, we will use the stoichiometry of the reaction that forms iron(III) oxide (Fe2O3). The balanced chemical equation for the formation of iron(III) oxide from iron and oxygen is:
4Fe + 3O2 → 2Fe2O3
From the equation, we see that 4 moles of iron react to form 2 moles of iron(III) oxide. Therefore, the stoichiometric mole ratio of Fe to Fe2O3 is 4:2, which simplifies to 2:1.To find the moles of Fe2O3 produced, we will set up a conversion factor that relates the moles of Fe used to the moles of Fe2O3 produced based on our balanced equation:
0.483 mol Fe × (1 mol Fe2O3 ÷ 2 mol Fe) = 0.242 mol Fe2O3
Thus, 27.0 grams of Fe would produce approximately 0.242 moles of Fe2O3 assuming oxygen is in excess.