Question

A solid mixture weighing 5 g containing lead nitrate and sodium nitrate was heated below 600

Answer 1

Let the mass of lead nitrate in the mixture be x g.Therefore mass of sodium nitrate in the mixture will be (5-x) g.

Molar mass of Pb(NO3)2 = 207 + (14 +16×3)×2 =​331g

Molar mass of PbO = 207 + 16 =​223 g

2 Pb(NO3)2 (s)→     2 PbO (s)      +   4 NO2 (g) + O2 (g)
2×331=662g          2×223=446g

662 g Pb(NO3)2  give residue = 446 g 
Therefore x g Pb(NO3)2 will give residue = 446662×x g 
                                                           = 0.674 x g

Molar mass of NaNO3 = 23 + (14 +16×3) =​ 85g

Molar mass of NaNO2 = 23 + 14 + 2×16 =  69 g

2 NaNO3 (s)→ 2NaNO2 (s)  + O2 (g)
2×85=170g      2×69=138g

170 g Pb(NO3)2 give residue = 138 g 
Therefore x g Pb(NO3)2 will give residue = 138170×(5−x )g 
                                                           = 0.812 (5-x)  g

Since upon heating a weight loss of 28% is observed thus weight of the residue obtained = 100-28 % = 72%
Thus weight of residue  = 72100×5
                                   = 3.6 g
∴ 0.674x + 0.812 (5-x)= 3.6
                        0.138  = 0.46
                         x        = 0.460.138
                              x  = 3.33 g 

Therefore Pb(NO3)2 in the mixture = x g = 3.33 g

Thus NaNO3 in the mixture = 5-x g
                                         =5 - 3.33g = 1.67 g

Answer 2

Answer:

Pb(NO3)2 in the mixture = x g = 3.33 g

Thus NaNO3 in the mixture = 5-x g

                                        =5 - 3.33g = 1.67 g