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

A solid mixture (5.0 g) consisting of lead nitrate and sodium nitrate

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