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

Loss in weight is due to formation of `NO_(2) and O_(2)` gasses `("weight loss"=(5)/(100)xx28=1.4g)`
`underset(xg)(Pb(NO_(3))_(2)) to underset((x-y)g)(PbO+NO_(2))+ubrace(NO_(2)+O_(2))_(yg)`
`underset(5-x)(NaNO_(3)) to underset((3.6-x+y))(NaNO_(2))+underset(1.4-y)(1//2O_(2))`
Number of moles of `Pb(NO_(3))_(2)`=Number of moles of PbO
`(x)/(331)=((x-y))/(223).........(i)`
`therefore M.w Pb(NO_(3))_(2)=331 M.w. PbO=223`
Number of moles of `NaNO_(3)`=Number of moles of `NaNO_(2)`
`(5-x)/(85)=(3.6-x+y)/(69)......(ii)`
Solving eqs. (i) and (ii) we get
`"Mass of "Pb(NO_(3))_(2)=X=3.3246g`
`"Mass of "NaNO_(3)=5-3.3246=1.6754g`