Question

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.

Answer 1

Correct Answer - `(1.7 g)`
Heating below `600^(@)C` converts `Pb(NO_(3))_(2)` into `PbO` but to `NaNO_(3)` into `NaNO_(2)` as
`{:(,Pb(NO_(3))_(2),overset(Delta)rarr,PbO(s)+,2NO_(2)uarr+ (1)/(2)O_(2)uarr),(MW:,330,,222,),(,NaNO_(3),overset(Delta)rarr,NaNO_(2)(s)+,(1)/(2)O_(2)uarr),(MW :,85,,69,):}`
Weight loss `= 5 xx (28)/(10) = 1.4 g`
`rArr` Weight of residue left `= 5-14 = 3.6 g`
Now, let the original mixture contain `x g` of `Pb(NO_(3))_(2)`.
`because 330 g Pb(NO_(3))_(2)` gives `222 gPbO`
`:. x g Pb(NO_(3))_(2)` will give `(222 x)/(330) g PbO`
Similarly, `85 g NaNO_(3)` gives `69 g NaNO_(2)`
`rArr (5-x)g NaNO_(3)` will give `(69(5-x))/(85)g NaNO_(2)`
`rArr` Residue: `(222 x)/(330) + (69(5-x))/(85) = 3.6 g`
Solving for x gives, `x = 3.3 g Pb(NO_(3))_(2)`
`rArr NaNO_(3) = 1.7 g`.