Question

A 350mg sample containing KClO3 was reduced and treated with exes AgNO3. The resulting Agcl weighed 185mg. Calculate the % KClO3 in the sample.

Answer 1

The scheme of conversation KClO3 into AgCl is:

$$\mathrm{KClO_3} \rightarrow \mathrm{KCl} \rightarrow \mathrm{AgCl}$$

Mole ratio of KClO3 and AgCl is 1: 1

Amount of AgCl is the same as KClO3 and it is:

$$n = 185 / 143 = 1,2934 \text{ mol}$$

Mass of KClO3 is Mw of it * amount = 122.5 * 1,2934 = 158,48 g

% of KClO3 is 158,48/350 * 100% = 45,28%