A solid mixture(5.000 g) consisting of lead nitrate and sodium nitrate was heated below 600°C until the weight of the residue was constant. If the loss in weight is 28%, find the amount of lead nitrate and sodium nitrate in the mixture.

A solid mixture(5.000 g) consisting of lead nitrate and sodium nitrate

1.	A solid mixture(5.000 g) consisting of lead nitrate and sodium nitrate was heated below 600°C until the weight of the residue was constant. If the loss in weight is 28%, find the amount of lead nitrate and sodium nitrate in the mixture.
Answer» Solution :Suppose, the given mixture CONTAINS x g of lead nitrate. `therefore` The amount of sodium nitrate in the mixture = `5.0 - x` g The corresponding equations are: `2Pb(NO_(3))_(2) to UNDERSET(2 xx 331.22 g)(2PbO) + 4NO_(2) uarr + underset(2 xx 223.2 g)(O_(2)) uarr` `2NaNO_(3) to underset(2.85 g)(2NaNO_(2)0 + underset(2 xx 69 g)(O_(2)) uarr` According to these equations, weight of PBO formed by heating x g of `Pb(NO_(3))_(2)` `=(2 xx 223.2)/(2 xx 331.22) xx x = 0.6739 g` and weight of `NaNO_2` obtained by heating `(5.0 -x) g` of `NaNO_(3)` `=(2 xx 69)/(2 xx 85) xx (5.0-x) = 0.8118(5.0 -x)`g of `NaNO_(3)` `therefore` Total weight of the residue = = 0.6739 x + 0.8118 (5.0 - x) g Since, the mixture loses 28% on heating, the residue obtained on heating 5.0 g of mixture `=(100-28)/100 xx 5.000 = 3.600 g` The amount of `Pb (NO_3)_2` in the mixture = 3.328 g and the amount of `NaNO_3` in the mixture = 5.0 - 3.328 = 1.672 g