Question

6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.2Al + 3 CuSO4 → Al2(SO4)3 + 3Cu What is the actual yield? What is the theoretical yield? What is the percent yield?

Answer 1

Actual yield = 6.78g

Theoretical yield = 13.85g

% yield = 48.95%

Explanations:

The balanced equation that results from the reaction of Almunimum and copper(II) sulfate is expressed as shown:

`2Al+3\text{CuSO}_4\rightarrow Al_2(SO_4)_3+3Cu`

From the balanced equation, we can see that 2 moles of Aluminum produced 3 moles of copper.

Get the number of moles of aluminum;

`\text{Moles}=\frac{Mass}{\text{Molar mass}}`

Mass of Al = 3.92g

Molar mass of Al = 26.98g/mol

`\begin{gathered} \text{Moles of Al=}(3.92)/(26.98) \\ \text{Moles of Al=}0.145\text{moles} \end{gathered}`

If 2 moles of Al produced 3 moles of Cu, then 0.145 moles of Al will produce;

`\begin{gathered} \text{Mol of Cu=}(0.145*3)/(2) \\ \text{Mole of Cu=}0.218\text{mol} \end{gathered}`

The calculated mass of copper will be the theoretical yield as shown:

`\begin{gathered} \text{Mass of Cu=Moles of Cu}* Molar\text{ mass of Cu} \\ \text{Mass of Cu=0.218}*63.55 \\ \text{Mass of Cu=}13.85\text{grams} \end{gathered}`

The theoretical yield is 13.85 grams

The actual yield is the given mass of copper which is 6.78 grams

Get the percentage yield

`\begin{gathered} \%\text{yield =}\frac{actual\text{ yield}}{theoretical\text{ yield}}*100 \\ \%\text{yield}=(6.78)/(13.85)*100 \\ \%\text{yield}=48.95\% \end{gathered}`

Hence the percent yield is 48.95%