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

Quizplus · Accepted Answer

First, we need to determine the limiting reactant. We can do this by converting both reactants to moles and comparing the mole ratios to the balanced equation.

11 mol Na2S × 1 mol CuSO4 / 1 mol Na2S = 11 mol CuSO4
11 mol CuSO4 is exactly the amount needed to react with all of the Na2S.

11 mol CuSO4 × 159.61 g / mol = 1755.7 g CuSO4

1755.7 g CuSO4 is exactly the amount needed to react with all of the Na2S.

11 mol Na2S × 78.06 g / mol = 858.66 g Na2S
858.66 g Na2S is in excess in this reaction.

To determine how much of the excess reactant remains, we need to calculate how much is used up in the reaction.

11 mol CuSO4 × 111.62 g CuS / mol = 1227.8 g CuS used
858.66 g Na2S - 1227.8 g CuS = -369.1 g Na2S

Since we cannot have a negative amount of reactant remaining, we know that all of the CuSO4 was used up and 369.1 g of Na2S remains.

1755.7 g CuSO4 - 1755.7 g CuSO4 used = 0 g CuSO4 remaining 
858.66 g Na2S - 369.1 g Na2S used = 489.56 g Na2S remaining

Therefore, the answer is C) 9.58 g of Na2S remain as excess reactant.

How Many Grams of the Excess Reactant Remain Assuming the Reaction