Final answer:
When 42.24 g of Fe reacts with oxygen, the stoichiometry calculation based on the balanced chemical equation yields approximately 60.36 g of Fe2O3 formed. Option a (60.39) is thus the closest answer.
Step-by-step explanation:
To determine how many grams of Fe2O3 are formed when 42.24 g of Fe reacts completely with oxygen, we use stoichiometry. Considering the balanced chemical equation:
4Fe(s) + 3O2(g) → 2Fe2O3(s)
We start by calculating the moles of Fe, using its molar mass (55.85 g/mol):
mole Fe = 42.24 g ÷ 55.85 g/mol = 0.756 mol Fe
Using the balanced equation, we find that the mole ratio of Fe to Fe2O3 is 4:2 or 2:1. Therefore, 0.756 mol of Fe will produce half as many moles of Fe2O3, which is 0.378 mol of Fe2O3. Then, we convert the moles of Fe2O3 to grams:
mass Fe2O3 = 0.378 mol × 159.69 g/mol = 60.36 g
Therefore, 42.24 g of Fe will produce approximately 60.36 g of Fe2O3, which when rounded to three significant figures is 60.4 g, making option a (60.39) the closest answer.