Final answer:
To calculate the grams of Fe₂O₃ formed from 11.8g of Fe burned in excess O₂, a stoichiometric calculation is done. The closest answer after conversion and stoichiometry, rounded to two significant figures, is 16.6 g of Fe₂O₃.
Step-by-step explanation:
To determine how many grams of Fe2O3 will form when 11.8g of Fe metal is burned in the presence of excess O2, we need to follow several steps that involve stoichiometry and molar mass calculations. First, we write the balanced chemical equation for the reaction:
4Fe + 3O2 → 2Fe2O3
Next, we find the molar mass of Fe and Fe2O3 which are 55.85 g/mol and 159.7 g/mol respectively. We then convert the mass of Fe to moles, and use the stoichiometry of the reaction to find the moles of Fe2O3 that will form:
11.8 g Fe × (1 mol Fe / 55.85 g Fe) = 0.211 moles of Fe
From the balanced equation, it is clear that 4 moles of Fe produce 2 moles of Fe2O3, so we set up a ratio:
(0.211 moles Fe) × (2 moles Fe2O3 / 4 moles Fe) = 0.1055 moles of Fe2O3
Finally, we convert the moles of Fe2O3 to grams using its molar mass:
0.1055 moles Fe2O3 × (159.7 g Fe2O3 / 1 mol Fe2O3) = 16.85 g of Fe2O3
Therefore, when 11.8g of Fe is burned, approximately 16.85g of Fe2O3 will form, but since this answer is not provided in the options and we should stick to two significant figures, the closest answer is (a) 16.6g.