A solid mixture 5 gram consists of lead nitrate and sodium nitrate was heated below ${600}^{0}C$until weight of residue was constant. If the loss in weight is 28%, find the amount of lead nitrate and sodium nitrate in mixture.

Following are thermal decomposition  processes.

$\underset{ag}{Pb{\left(N{O}_3\right)}_2}\to PbO+2N{O}_2\uparrow +\frac{1}{2}{O}_2\uparrow$

$\underset{bg}{NaN{O}_3}\to NaN{O}_2+\frac{1}{2}{O}_2\uparrow$

$\therefore a+b=5\mathrm{.....}\left(1\right)$

The loss in weight for 5 gram mixture  $=5\times \frac{28}{100}=1.4gram$

Residue left = 5 – 1.4 = 3.6 gram

The residue contains $PbO+NaN{O}_2$

$\because 331gramPb{\left(N{O}_3\right)}_{2}gives=223gPbO$

$\therefore agPb{\left(N{O}_3\right)}_{2}gives=\frac{223\times a}{331}gPbO$

$\because 85gramNaN{O}_{3}gives=69gramNaN{O}_2$

$\therefore bgNaN{O}_{3}gives=\frac{69\times b}{85}gramNaN{O}_2$

$\therefore \frac{223\times a}{331}+\frac{69\times b}{85}=3.6$                             …..(2)

Solving equations (1) and (2),

$\therefore a=3.32gram,b=1.68gram$