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

Mass of solid mixture = 5.0 g
Let the mass of lead nitrate = x g
`:.` Mass of sodium nitrate `= (5-x)g`
Loss in mass of mixture `= (5xx28)/(100)=1.4g`
`:.` Mass of the residue `= 5 - 4 = 3.6 g`
Calculation of the mass of the residue from lead nitrate.
`underset(underset("= 663 g")(2(207+28+96)))(2Pb(NO_(3))_(2))rarrunderset(underset(=446g)(2(207+16)))(2PbO)+4NO_(2)+O_(2)`
662 g of `Pb(NO_(3))_(2)` form residue (PbO) = 446 g
`:. x` g of `Pb(NO_(3))_(2)` form residue `(PbO)=(446)/(662) xx xg`
Step II. Calculation of the mass of the residue from sodium nitrate
`underset(underset(=170g)(2(23+14+48)))(2NaNO_(3))rarrunderset(underset(=138g)(2(23+14+32)))(2NaNO_(2)+O_(2))`
170g of `NaNO_(3)` form residue `(NaNO_(2))=138g`
`:. (5-x)g` of `NaNO_(3)` form residue `(NaNO_(2))=(138)/(170) xx (5-x)g`
Step III. Calculation of the amount of lead nitrate and sodium nitrate
Total mass of PbO and `NaNO_(3)=[(446)/(662)x+(138)/(170)(5-x)]g`
Total mass of the residue actually left = 3.6 g
`:. (446)/(662)x+(138)/(170)(5-x)=3.6`
`0.674x+0.812(5-x)=3.6`
or `0.674x+4.06 -0.812x = 3.6`
or `0.674x-0.812x = 3.6 - 4.06`
`-0.13x = - 0.46`
`:. x = (0.46)/(0.138)=3.38g`
Thus, Mass of `Pb(NO_(3))_(2)=3.38g`
Mass of `NaNO_(3)=5 - 3.38 = 1.62g`.