A solid mixture (5g) consistingof lead nitrate andsodiumnitrate was heated below 600^(@) Cuntil the weightof the residueis constant. If the loss in weight is 28%, findthe amountof the leadnitrate ad sodium nitrate in themixture.

A solid mixture (5g) consistingof lead nitrate andsodiumnitrate was he

1.	A solid mixture (5g) consistingof lead nitrate andsodiumnitrate was heated below 600^(@) Cuntil the weightof the residueis constant. If the loss in weight is 28%, findthe amountof the leadnitrate ad sodium nitrate in themixture.
Answer» SOLUTION :The LOSS in weight `(28% of 5 g)`, i.e, 1 . 4 g, is due to the formation of thegases `NO_(2) and O_(2)`whichescape out `{:(PB (NO_(3))_(2) ""to " " Pb O ""+"" ubrace(NO_(2)+O_(2))),("x g (say)(x - y) gy g (say) "),(" "NaNO_(3)" " to" " NaNO_(2)+ O_(2)),("(5 - x)(3 . 6 - x + y) ( 1 . 54 - y)" ):}` Applying POAC for Pband 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]` fromwhich, 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