Question

The moles of chlorine liberated by 79g of potassium permanganate on treatment with potassium chloride in a medium of dilute sulfuric acid:
a) 0.5
b)0.75
c)1
d)1.25

Answer 1

Answer to Question #53100 – Chemistry – Inorganic Chemistry

The moles of chlorine liberated by 79g of potassium permanganate on treatment with potassium chloride in a medium of dilute sulfuric acid:

a) 0.5

b) 0.75

c) 1

d) 1.25

Solution:

$$2 \mathrm{KMnO_4} + 10 \mathrm{KCl} + 8 \mathrm{H_2SO_4} \rightarrow 6 \mathrm{K_2SO_4} + 2 \mathrm{MnSO_4} + 5 \mathrm{Cl_2} + 8 \mathrm{H_2O}$$

Potassium permanganate, $\mathrm{Mr} = 158\ \mathrm{g/mol}$

$$n_{\mathrm{KMnO_4}} = \frac{79\ \mathrm{g}}{158\ \mathrm{g/mol}} = 0.5\ \mathrm{mol}$$ $$n_{\mathrm{Cl_2}} = 5 \times \frac{n_{\mathrm{KMnO_4}}}{2} = \frac{5 \times 0.5\ \mathrm{mol}}{2} = 1.25\ \mathrm{mol}$$

Answer: d) 1.25

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