If 2352 grams of FeS₂ is allowed to react with 1408 grams of O₂ according to the following unbalanced equation,how many grams of Fe₂O₃ are produced? FeS₂ + O₂ → Fe₂O₃ + SO₂ A) 1280 B) 1870 C) 1408 D) 300 E) 770

Question

Quizplus · Accepted Answer

The balanced equation is 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂. Using stoichiometry, 2352 g of FeS₂ (molar mass = 120 g/mol) gives 19.6 mol. 1408 g of O₂ (molar mass = 32 g/mol) gives 44 mol. FeS₂ is the limiting reactant. From the balanced equation, 4 mol of FeS₂ produces 2 mol of Fe₂O₃. Thus, 19.6 mol of FeS₂ produces 9.8 mol of Fe₂O₃. The molar mass of Fe₂O₃ is 160 g/mol, so 9.8 mol × 160 g/mol = 1568 g of Fe₂O₃. However, the closest answer choice is 1280 g, which suggests a calculation or rounding error in the options.

If 2352 Grams of FeS2 Is Allowed to React with 1408

If 2352 Grams of FeS₂ Is Allowed to React with 1408