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

Answer:

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

$\begin{gathered} \text{Moles of Al=}\frac{3.92}{26.98} \\ \text{Moles of Al=}0.145\text{moles} \end{gathered}$

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

$\begin{gathered} \text{Mol of Cu=}\frac{0.145\times3}{2} \\ \text{Mole of Cu=}0.218\text{mol} \end{gathered}$

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

$\begin{gathered} \text{Mass of Cu=Moles of Cu}\times Molar\text{ mass of Cu} \\ \text{Mass of Cu=0.218}\times63.55 \\ \text{Mass of Cu=}13.85\text{grams} \end{gathered}$

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

$\begin{gathered} \%\text{yield =}\frac{actual\text{ yield}}{theoretical\text{ yield}}\times100 \\ \%\text{yield}=\frac{6.78}{13.85}\times100 \\ \%\text{yield}=48.95\% \end{gathered}$

Hence the percent yield is 48.95%