Question

If 5.00 grams of aluminum react with an excess of copper (II) sulfate and the percentage yield is 63.4%, what is the mass of the copper produced? The other product is aluminum sulfate.

Answer 1

Answer : The mass of copper produced will be, 11.796 grams

Explanation : Given,

Mass of
`Al` = 5 g

Molar mass of
`Al` = 26.98 g/mole

Molar mass of
`Cu` = 63.66 g/mole

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

`\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}=(5g)/(26.98g/mole)=0.185moles`

Now we have to calculate the moles of
`Cu`.

The balanced chemical reaction is,

`2Al+3CuSO_4\rightarrow Al_2(SO_4)_3+3Cu`

From the balanced reaction we conclude that

As, 2 moles of
`Al` react to give 3 moles of
`Cu`

So, 0.185 moles of
`Al` react to give
`(3)/(2)* 0.185=0.2775` 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=(0.2775mole)* (63.66g/mole)=17.66g`

The theoretical yield of Cu = 17.66 grams

Now we have to calculate the actual yield of Cu.

`\%\text{ yield of }Cu=\frac{\text{Actual yield of }Cu}{\text{Theoretical yield of }Cu}* 100`

Now put all the given values in this formula, we get the actual yield of Cu.

`63.4=\frac{\text{Actual yield of }Cu}{17.66}* 100`

`\text{Actual yield of }Cu=11.796g`

Therefore, the mass of copper produced will be, 11.796 grams