Question

A sample of $CuS{O}_4.5H_{2}O$ having 40 % purity by moles undergoes the following sequence of reactions in a flask having large amounts of $KCN$.
$CuS{O}_4.5H_{2}O\to CuS{O}_4+5H_{2}O\text{(Yield = 100\%)}$
$CuS{O}_4+2KCN\to Cu(CN{)}_2+K_{2}S{O}_4\text{(Yield = 80\%)}$
$2Cu(CN{)}_2\to 2CuCN+(CN{)}_2\text{(Yield = 60\%)}$
$2CuCN+6KCN\to 2K_3\left[Cu\right(CN{)}_4]\text{(Yield = 50\%)}$
Find the moles of impure $CuS{O}_4.5H_{2}O$ required for producing 28.45 g of the complex.
(Molar mass of Cu = 63.5 g/mol; molar mass of K = 39 g/mol)

• A. 1.15 mol
• B. 1.04 mol
• C. 4.12 mol
• D. 3.36 mol

Answer 1

The correct option is B 1.04 mol

$CuS{O}_4.5H_{2}O\to CuS{O}_4+5H_{2}O\text{(1) (Yield = 100 \%)}$
$CuS{O}_4+2KCN\to Cu(CN{)}_2+K_{2}S{O}_4\text{(2) (Yield = 80 \%)}$
$2Cu(CN{)}_2\to 2CuCN+(CN{)}_2\text{(3) (Yield = 60 \%)}$ $2CuCN+6KCN\to 2K_3\left[Cu\right(CN{)}_4]\text{(4) (Yield = 50 \%)}$
Number of moles of $K_3\left[Cu\right(CN{)}_4]=\frac{28.45}{284.5}=0.1$ mol
In reaction (4),
2 moles of $K_3\left[Cu\right(CN{)}_4]$ is produced by 2 moles of $CuCN$
0.1 mole will be produced by 0.1 mole of $CuCN$
Yield of the reaction is given as 50 %
Actual moles required = $\frac{0.1}{0.5}$ = 0.2 moles
In reaction (3),
2 mole of $CuCN$ is produced by 2 mole of $Cu(CN{)}_2$
0.2 mole will be produced by $=0.2$ mole
Yield of the reaction is given as 60 %
Actual moles required = $\frac{0.2}{0.6}$ = 0.33 mol
In reaction (2),
1 mole of $Cu(CN{)}_2$ is produced by 1 mole of $CuS{O}_4$
0.33 mole will be produced by 0.33 mole of $CuS{O}_4$
Yield of the reaction is given as 80 %
Actual moles required = $\frac{0.33}{0.8}$ = 0.4125 mol
In reaction (1),
1 mole of $CuS{O}_4.5H_{2}O$ will be required to produce 1 mole of $CuS{O}_4$
0.4125 moles of $CuS{O}_4.5H_{2}O$ will be required to produce 0.4125 mole of $CuS{O}_4$ since yield is 100 %.
As the initial sample is 40 % pure by moles
Actual moles of $CuS{O}_4.5H_{2}O=\frac{0.4125}{40}\times 100=1.04$ mol