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

`Pb(NO_(3))_(2) = 3.3g, NaNO_(3) = 1.7 g`
Heating below 600°C converts `Pb(NO_(3))_(2)` into PbO and `NaNO_(3)` into `NaNO_(2)` as
`MW : underset(330)(Pb(NO_(3))_(2)) overset(Delta)rarr underset(222)(PbO(s)) + 2NO_(2) uarr +(1)/(2)O_(2) uarr , MW : underset(85)(NaNO_(3)) overset(Delta)rarr underset(69)(NaNO_(2))(s) + (1)/(2)O_(2)uarr`
Weight loss `= 5 xx (28)/(100) = 1.4 g rArr` Weight of residue left = 5 - 1.4 = 3.6 g
Now, 330 g `Pb(NO_(3))_(2)` ives 222 g PbO `:. xg Pb(NO_(3))_(2)` will give `(222x)/(330)g PbO`
Similarly, 85 g `NaNO_(3)` gives 69 g `NaNO_(2) rArr (5 - x)g NaNO_(3)` will give `(69(5-x))/(85) g NaNO_(2)`
`rArr` Residue : `(222 x)/(330) + (69(5-x))/(85) = 3.6 g`
Solving for x gives, `x = 3.3 g Pb (NO_(3))_(2) rArr NaNO_(3) = 1.7 g`