Question

What is the theoretical yield in grams of CuS for the following reaction given that you start with 15.5 g of Na2S and 12.1 g CuSO4? Reaction: Na2S + CuSO4 → Na2SO4 + CuS

Answer 1

Answer : The theoretical yield of CuS is, 7.26 grams.

Solution : Given,

Mass of $Na_2S$ = 15.5 g

Mass of $CuSO_4$ = 12.1 g

Molar mass of $Na_2S$ = 78 g/mole

Molar mass of $CuSO_4$ = 160 g/mole

Molar mass of CuS = 96 g/mole

First we have to calculate the moles of $Na_2S$ and $CuSO_4$.

$\text{ Moles of }Na_2S=\frac{\text{ Mass of }Na_2S}{\text{ Molar mass of }Na_2S}=\frac{15.5g}{78g/mole}=0.199moles$

$\text{ Moles of }CuSO_4=\frac{\text{ Mass of }CuSO_4}{\text{ Molar mass of }CuSO_4}=\frac{12.1g}{160g/mole}=0.0756moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Na_2S+CuSO_4\rightarrow Na_2SO_4+CuS$

From the balanced reaction we conclude that

As, 1 mole of $CuSO_4$ react with 1 mole of $Na_2S$

So, 0.0756 moles of $CuSO_4$ react with 0.0756 moles of $Na_2S$

From this we conclude that, $Na_2S$ is an excess reagent because the given moles are greater than the required moles and $CuSO_4$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CuS$

From the reaction, we conclude that

As, 1 mole of $CuSO_4$ react to give 1 mole of $CuS$

So, 0.0756 moles of $CuSO_4$ react to give 0.0756 moles of $CuS$

Now we have to calculate the mass of $CuS$

$\text{ Mass of }CuS=\text{ Moles of }CuS\times \text{ Molar mass of }CuS$

$\text{ Mass of }CuS=(0.0756moles)\times (96g/mole)=7.26g$

Thus, the theoretical yield of CuS is, 7.26 grams.