How many grams of the excess reactant remain assuming the reaction goes to completion and that you start with 15.5 g of Na₂S and 12.1 g CuSO₄?
Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

Question

How many grams of the excess reactant remain assuming the reaction goes to completion and that you start with 15.5 g of Na₂S and 12.1 g CuSO₄? Reaction: Na₂S + CuSO₄→ Na₂SO₄ + CuS

Quizplus · Accepted Answer

First, we need to determine the limiting reactant. To do this, we can use the stoichiometry of the reaction to convert both reactants to the same product (CuS). 15.5 g Na₂S x (1 mol Na₂S/117.16 g Na₂S) x (1 mol CuS/1 mol Na₂S) x (63.55 g CuS/1 mol CuS) = 7.64 g CuS 12.1 g CuSO₄S x (1 mol CuSO₄S/159.61 g CuSO₄S) x (1 mol CuS/1 mol CuSO₄S) x (63.55 g CuS/1 mol CuS) = 5.94 g CuS We can see that CuSO₄S is the limiting reactant. Therefore, all of the Na₂S will react completely, and we need to determine how much of the excess CuSO₄S is left over. 5.94 g CuS - 5.94 g CuSO₄S = 0 g CuSO₄S remaining Therefore, the excess reactant is: 12.1 g CuSO₄S - 5.94 g CuSO₄S = 6.16 g CuSO₄S To find the answer with the correct significant figures, we round to two decimal places, giving us 9.58 g CuSO₄S remaining. The answer is C.

How Many Grams of the Excess Reactant Remain Assuming the Reaction