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}=(15.5g)/(78g/mole)=0.199moles`

`\text{ Moles of }CuSO_4=\frac{\text{ Mass of }CuSO_4}{\text{ Molar mass of }CuSO_4}=(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* \text{ Molar mass of }CuS`

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

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