# 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 wei

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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.

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Let the amount of NaNO_(3) in the mixture be =xg
The amount of Pb(NO_(3))_(2) in the mixture=(5.0-x)g
underset(2xx85g)(2NaNO_(3)) overset("Heat")to2NaNO_(2)+underset(32g)(O_(2))
underset(662g)underset((2xx331g))(2Pb(NO_(3))_(2))overset("Heat")to ubrace(2PbO+4NO_(2)+O_(2))_(216g)
170g of NaNO_(3) evolve oxygen=32g
x g of NaNO_(3) evolve oxygen =(32)/(170)xx xg
662g of Pb(NO_(3))_(2) evolve gases=216g
(50-x)g "of"Pb(NO_(3))_(2) evolve gases=(216)/(662)xx(50-x)g
Total loss=(32)/(170)xx x+(216)/(662)xx(5.0-x)
Loss given in the problem=(28)/(100)xx5=1.4g
therefore (32)/(170)x+(216)/(662)(5.0-x)=1.4
On solving x=1.676g
Thus, Mass of NaNO_(3)=1.676g
Mass of Pb(NO_(3))_(2)=(5.0-1.676g)g=3.324g