Question

2.80g copper chloride and 0.400 g aluminum are mixed together and allowed to react, what is the theoretical yield of copper?

Answer 1

Answer : The theoretical yield of copper is 1.98 grams.

Solution : Given,

Mass of
`CuCl_2` = 2.80 g

Mass of
`Al` = 0.400 g

Molar mass of
`CuCl_2` = 134.45 g/mole

Molar mass of
`Al` = 26.9 g/mole

Molar mass of Cu = 63.5 g/mole

First we have to calculate the moles of
`CuCl_2` and
`Al`.

`\text{ Moles of }CuCl_2=\frac{\text{ Mass of }CuCl_2}{\text{ Molar mass of }CuCl_2}=(2.80g)/(134.45g/mole)=0.0208moles`

`\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=(0.400g)/(26.9g/mole)=0.0149moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`3CuCl_2+2Al\rightarrow 3Cu+2AlCl_3`

From the balanced reaction we conclude that

As, 3 mole of
`CuCl_2` react with 2 mole of
`Al`

So, 0.0208 moles of
`CuCl_2` react with
`(0.0208)/(3)* 2=0.0139` moles of
`Al`

From this we conclude that,
`Al` is an excess reagent because the given moles are greater than the required moles and
`CuCl_2` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`Cu`

From the reaction, we conclude that

As, 2 mole of
`CuCl_2` react to give 3 mole of
`Cu`

So, 0.0208 moles of
`CuCl_2` react to give
`(0.0208)/(2)* 3=0.0312` moles of
`Cu`

Now we have to calculate the mass of
`Cu`

`\text{ Mass of }Cu=\text{ Moles of }Cu* \text{ Molar mass of }Cu`

`\text{ Mass of }Cu=(0.0312moles)* (63.5g/mole)=1.98g`

Therefore, the theoretical yield of copper is 1.98 grams.