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

`underset("ag")Pb(NO_(3))_(2) rarr PbO + 2 NO_(2) uarr + 1/2 O_(2) uarr`
`underset("bg")NaNO_(3) rarr NaNO_(2) + 1/2 O_(2) uarr`
a + b = 5 g ........ (i)
The loss in weight for 5 g mixture = 5 xx 28/100 = 1.4 g
Residue left = 5 - 1.4 = 3.6 g
The residue contains `PbO + NaNO_(2)`
`331 g Pb(NO_(3))_(2)` gives = 223 g PbO
`a g Pb(No_(3))_(2)` gives = `223 xx a/332` g PbO
Similarly,
85 g `NaNO_(3)` gives = 69 g `NaNO_(3)`
`b g NaNO_(3) gives = 69 xx b/85 g NaNO_(2)`
`223xxa/331 + 69xxb/85 = 3.6 ` ........ (ii)
Solving equation, (i) and (ii) a = 3.32 g and b = 1.68 g