Question

A solid mixture (5.0 g) consisting of lead nitrate and sodium nitrate was heated below `600^(@)C` untill the weight of the residue was constant . If the loss in weight is 20% find the amount of lead nitrate and sodium nitrate in the mixture.

Answer 1

`underset(ag)(Pb(NO_(3))_(2))to PbO +2NO_(2) uarr + (1)/(2)O_(2) uarr`
`underset(bg)(NaNO_(3)) to NaNO_(2)+(1)/(2)O_(2)uarr`
`therefore a+b=5" "…(1)`
The loss in weight for 5 g mixture `=5xx(28)/(100)=1.4 g`
`therefore` Residue left=5-1.4=3.6 g
The residue contain `PbO+NaNO_(2)`
`because` 331 g `(Pb(NO_(3))_(2)` gives=223 g PbO
`therefore agPb(NO_(3))_(2)` gives `=(223xxa)/(332)gPbO`
Similarly,
`therefore `85 g `NaNO_(3)` gives =`69 NaNO_(3)`
`therefore` bg `NaHO_(3)` gives `=(69xxb)/(85) g NaNO_(3)`
Solving equation , (1) and (2)
a=3.32 g and b=1.68 g