Final answer:
To find how many grams of Fe2O3 are formed from 42.24 g of Fe, one must perform stoichiometric calculations. The mass of iron is converted to moles, which are then used to find the moles of Fe2O3. These moles are converted to grams, yielding an answer of approximately 60.36 g, which corresponds with option A) 60.39 g.
Step-by-step explanation:
To determine how many grams of Fe2O3 are formed when 42.24 g of Fe reacts completely with oxygen, it is necessary to use stoichiometric calculations based on the balanced chemical equation. The balanced equation for the formation of iron(III) oxide from iron and oxygen is: 4Fe + 3O2 → 2Fe2O3.
First, we convert the given mass of iron to moles using the molar mass of iron (55.85 g/mol). Then, we use the stoichiometric ratios from the equation to find the moles of Fe2O3 that can be formed. Finally, we convert moles of Fe2O3 to grams using its molar mass (159.69 g/mol). Following these steps, the calculation is as follows:
Mass of Fe = 42.24 g
Molar mass of Fe = 55.85 g/mol
Moles of Fe = 42.24 g ÷ 55.85 g/mol = 0.756 mol of Fe
Now, using the mole ratio of Fe to Fe2O3 from the balanced equation, which is 4:2 or 2:1, we find:
Moles of Fe2O3 = 0.756 mol Fe × (1 mol Fe2O3/2 mol Fe) = 0.378 mol of Fe2O3
Finally, we use the molar mass of Fe2O3 to find the mass:
Mass of Fe2O3 = 0.378 mol × 159.69 g/mol = 60.36 g (rounded to three significant figures)
The answer to how many grams of Fe2O3 are formed is approximately 60.36 g which corresponds to option A) 60.39 g.