A solid mixture `5 g` consists of lead nitrate and sodium nitrate was heated below `600^(@)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.

Explanation

Let the amount of `NaNO_(3)` in the mixture
`underset(2(23 + 14 + 3 xx 16)= 170 g)(2NaNO_(3)) to 2NaNO_(2) + underset(2 xx 16= 32)(O_(2)) uarr`
`underset(662 g)(2Pb(NO_(2))_(2)) to 2PbO + underset(32)(O_(2) uarr) + underset(184 g)(4NO_(2)) uarr`
170 g of `NaNO_(3)` give `O_(2) = 32`g
x g of `NaNO_(3)` give `O_(2) = 32/170 xx x g`
Similarly,
662 g of `Pb(NO_(3))_(2)` give `O_(2)` and NO = 216 g
(5-x) g of `Pb(NO_(3))_(2)` give gases `=(216)/662 xx (5-x)g`
Total loss on heating `=(32 x)/170 + (216)/662 (5-x)`
Actual loss on heating = 28% of 5 g `=(28 xx 5)/100 = 1.4`g
`therefore (32 x)/170 + 216/662 (5-x) = 1.4`
Solving for x, we get
`therefore` Wt. of `NaNO_(3) = 1.676`g
Wt. of `Pb(NO_(3))_(2) = 5-1.676= 3.324`g.