[A] A solid mixture weighing `5.00g` containing lead nitrate and sodium nitrate was heated below `600^(@)C` until the mass of the residue was constant. If the loss of mass is `28%` find the mass of lead nitrate and sodium nitrate in the mixture.

Explanation

`{:(" "2Pb(NO_(3))_(2),overset(Delta)rarr," "2PbO,+4NO_(2)uarr+O_(2)uarr),(2xx331=662 g,,2xx233=446g,),(" "2 NaNO_(3),overset(Delta)rarr," "2NaNO_(2),+O_(2)uarr),(2xx85=170 g,,2xx69=138 g,):}`
Suppose `Pb(NO_(3))_(2)` in the mixture = x g
Then `NaNO_(3)` in the mixture = (5 - x) g
`662 g Pb(NO_(3))_(2)` give residue = 446 g
`therefore xg Pb(NO_(3))_(2)` will give reside `=(446)/(662)xx x g=0.674 x g`
`170 g NaNO_(3)` give residue = 138 g
`therefore (5-x)g NaNO_(3)` will give reside `=(138)/(170)xx(5-x)g=0.812(5-x)g`
Actual residue obtained = Mass taken - Mass lost `=5-(28)/(100)xx5 g=3.6 g`
Thus, total reside `=0.674 x + 0.812(5-x)=3.6` or `0.138 x =0.46` or `x=3.33 g`
i.e., `Pb(NO_(3))` in the mixture = 3.33 g
`NaNO_(3)` in the mixture `=5-3.33 = 1.67 g` .