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 $Al$.

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

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 $\frac{3}{2}\times 0.185=0.2775$ moles of $Cu$

Now we have to calculate the mass of $Cu$.

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

$\text{Mass of }Cu=(0.2775mole)\times (63.66g/mole)=17.66g$

The theoretical yield of Cu = 17.66 grams

Now we have to calculate the actual yield of Cu.

$\%\text{ yield of }Cu=\frac{\text{Actual yield of }Cu}{\text{Theoretical yield of }Cu}\times 100$

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

$63.4=\frac{\text{Actual yield of }Cu}{17.66}\times 100$

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

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