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 mass of `Pb(NO_(3))_(2)` be a gram.
`Pb(NO_(3))_(2)toPbO+2NO_(2)uarr+(1)/(2)O_(2)uarr`
for `331 g Pb (NO_(3))_(2)`, loss in weight`=108g`
For a gram `Pb(NO_(3))_(2)`, loss in weight`=(108a)/(331)`
Let the mass of `NaNO_(3)` be b gram.
For 331 g `NaNO_(3)`, loss in weight`=16g`
For b gram `NaNO_(3)`, loss in weight`=(16a)/(85)`
Loss in weight`=(5)/(100)xx28g`
`therefore(108a)/(331)+(16b)/(85)=(5)/(100)xx28` ..(i)
`a+b=5`
On solving equations (i) and (ii) we get,
`a=3.32g,b=1.68g`