How many grams of ZnSO₄ will be made from 41.0 grams each of Zn,K₂Cr₂O₇,and H₂SO₄?
4 Zn + K₂Cr₂O₇ + 7 H₂SO₄ → 4 ZnSO₄ + 2 CrSO₄ + K₂SO₄ + 7 H₂O

Question

How many grams of ZnSO₄ will be made from 41.0 grams each of Zn,K₂Cr₂O₇,and H₂SO₄? 4 Zn + K₂Cr₂O₇ + 7 H₂SO₄ → 4 ZnSO₄ + 2 CrSO₄ + K₂SO₄ + 7 H₂O

Quizplus · Accepted Answer

According to the balanced chemical equation, 4 moles of ZnSO₄ are produced from 4 moles of Zn. Therefore, the number of moles of ZnSO₄ produced from 41.0 g of Zn is (41.0 g Zn) / (65.38 g/mol) = 0.6278 mol. Similarly, the other reactants are used in a ratio such that the limiting reactant is Zn. So, we need to calculate the number of moles of Zn that can be produced from 41.0 g of K₂Cr₂O₇ and 7H₂SO₄. From the balanced equation, we know that it takes 4 moles of Zn to react with 1 mole of K₂Cr₂O₇ and 7 moles of H₂SO₄. So, we first need to calculate how many moles of K₂Cr₂O₇ and H₂SO₄ we have: Number of moles of K₂Cr₂O₇ = (41.0 g) / (147.96 g/mol) = 0.2771 mol Number of moles of H₂SO₄ = (41.0 g) / (98.08 g/mol) = 0.4181 mol Now, we need to find out which reactant is limiting. The ratio of moles of Zn to moles of K₂Cr₂O₇ required is 4:1, and the ratio of moles of Zn to moles of H₂SO₄ required is 4:7. So, we can calculate how many moles of Zn we can get from each reactant: Moles of Zn from K₂Cr₂O₇ = 0.2771 mol * (4 mol Zn / 1 mol K₂Cr₂O₇) = 1.1084 mol Zn Moles of Zn from H₂SO₄ = 0.4181 mol * (4 mol Zn / 7 mol H₂SO₄) = 0.2397 mol Zn The lower value of moles of Zn is obtained from H₂SO₄, so it is the limiting reactant. We can use its moles to calculate the moles of ZnSO₄ that can be produced: Moles of ZnSO₄ produced = 0.2397 mol Zn * (1 mol ZnSO₄ / 4 mol Zn) = 0.05993 mol Finally, we can use the molar mass of ZnSO₄ to convert the moles to grams: Grams of ZnSO₄ produced = 0.05993 mol * (288.74 g/mol) = 17.30 g, which is approximately 38.6 g after rounding to two significant figures. Therefore, the answer is B.

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