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 be =xg
The amount of `Pb(NO_(3))_(2)` in the mixture=(5.0-x)g
`underset(2xx85g)(2NaNO_(3)) overset("Heat")to2NaNO_(2)+underset(32g)(O_(2))`
`underset(662g)underset((2xx331g))(2Pb(NO_(3))_(2))overset("Heat")to ubrace(2PbO+4NO_(2)+O_(2))_(216g)`
170g of `NaNO_(3)` evolve oxygen=32g
x g of `NaNO_(3)` evolve oxygen =`(32)/(170)xx xg`
662g of `Pb(NO_(3))_(2)` evolve gases=216g
`(50-x)g "of"Pb(NO_(3))_(2)` evolve gases=`(216)/(662)xx(50-x)g`
Total loss=`(32)/(170)xx x+(216)/(662)xx(5.0-x)`
Loss given in the problem=`(28)/(100)xx5=1.4g`
`therefore (32)/(170)x+(216)/(662)(5.0-x)=1.4`
On solving x=1.676g
Thus, Mass of `NaNO_(3)=1.676g`
Mass of `Pb(NO_(3))_(2)=(5.0-1.676g)g=3.324g`