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

The loss in weight `(28% of 5 g)`, i.e, 1 . 4 g, is due to the formation of the gases `NO_(2) and O_(2)` which escape out
`{:(Pb (NO_(3))_(2) " "to " " Pb O " "+" " ubrace(NO_(2)+O_(2))),("x g (say) (x - y) g y g (say) "),(" "NaNO_(3)" " to" " NaNO_(2)+ O_(2)),(" (5 - x) (3 . 6 - x + y) ( 1 . 54 - y)" ):}`
Applying POAC for Pb and Na atoms, we get respectively,
moles of Pb `(NO_(3))_(2) ` = moles of PbO
`(x)/( 331) = (x - y)/( 223) " "` {Pb`(NO_(3))_(2)` = 331, PbO = 223}
and moles of Na`NO_(3)` = moles of Na`NO_(2) [NaNO_(3) = 85 , NaNO_(2) = 69]`
from which, we get,
x = 3 . 3246 g
`(5 - x)/( 85) = (3 . 6 - x + y)/( 69)b .(NaNO_(3) = 85, NaNO_(2) = 69)`
Thus, wt. of Pb `(NO_(3))_(2) = 3.3246 g`
wt. of `NaNO_(3) = 5 - 3 . 3246 = 1.6754` g