Question

5.33 gram of $CrCl_3 \cdot 6H_2O$ (1:3 electrolyte) is passed through cation exchanger. The resulting solution is then treated with an excess of $AgNO_3$, leading to formation of 8.61 gran of precipitate. Calculate :
$\frac{\text{number of moles of complex reacted}}{\text{number of moles of AgCl precipitated}} \times 100$

Answer 1

Stoichiometry involving coordination compounds and precipitation reactions.
1. Find moles of complex:
Molar mass of $CrCl_3 \cdot 6H_2O = 52 + (3 \times 35.5) + (6 \times 18) = 52 + 106.5 + 108 = 266.5 \text{ g/mol}$.
Moles of complex = $\frac{5.33}{266.5} = 0.02 \text{ moles}$.

2. Find moles of precipitate ($AgCl$):
Molar mass of $AgCl = 107.9 + 35.5 = 143.4 \text{ g/mol}$.
Moles of $AgCl = \frac{8.61}{143.5} \approx 0.06 \text{ moles}$.

3. Calculate the requested ratio:
Ratio = $\frac{\text{moles of complex}}{\text{moles of } AgCl} \times 100$
Ratio = $\frac{0.02}{0.06} \times 100 = \frac{1}{3} \times 100 = 33.33$

The answer is 33.