A solid mixture weighing 5.00 g containing lead nitrate and sodium nitrate was heated below 600^(@)C until the mass of the mixture. (At wts. of Pb = 207, Na = 23, N = 14, O = 16)

A solid mixture weighing 5.00 g containing lead nitrate and sodium nit

1.	A solid mixture weighing 5.00 g containing lead nitrate and sodium nitrate was heated below 600^(@)C until the mass of the mixture. (At wts. of Pb = 207, Na = 23, N = 14, O = 16)
Answer» Solution :`{:(" "2Pb(NO_(3))_(2),overset(Delta)rarr,""2PbO,+4NO_(2)uarr+O_(2)uarr),(2xx331=662 g,,2xx233=446g,),(" "2 NaNO_(3),overset(Delta)rarr," "2NaNO_(2),+O_(2)uarr),(2xx85=170 g,,2xx69=138 g,):}` Suppose `Pb(NO_(3))_(2)` in the MIXTURE = X g Then `NaNO_(3)` in the mixture = (5 - x) g `662 g Pb(NO_(3))_(2)` give residue = 446 g `therefore xg Pb(NO_(3))_(2)` will give reside `=(446)/(662)xx x g=0.674 x g` `170 g NaNO_(3)` give residue = 138 g `therefore (5-x)g NaNO_(3)` will give reside `=(138)/(170)xx(5-x)g=0.812(5-x)g` Actual residue obtained = Mass taken - Mass lost `=5-(28)/(100)xx5 g=3.6 g` Thus, total reside `=0.674 x + 0.812(5-x)=3.6` or `0.138 x =0.46` or `x=3.33 g` i.e., `Pb(NO_(3))` in the mixture = 3.33 g `NaNO_(3)` in the mixture `=5-3.33 = 1.67 g` .