Question

A solid mixture (5.0 g) consisting of lead nitrate and sodium nitrate was heated below 600C , until the weight of the residue was constant . If the loss in weight is 28%, calculate the amount of lead nitrate and sodium nitrate in the mixture

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$\times$3)$\times$2 =​331g

Molar mass of PbO = 207 + 16 =​223 g

2 Pb(NO₃)₂ (s)$\to$ 2 PbO (s) + 4 NO₂ (g) + O₂ (g)
2$\times$331=662g 2$\times$223=446g

662 g Pb(NO₃)₂ give residue = 446 g
Therefore x g Pb(NO₃)₂ will give residue = $\frac{446}{662}\times \mathrm{x}\mathrm{g}$
= 0.674 x g

Molar mass of NaNO₃ = 23 + (14 +16$\times$3) =​ 85g

Molar mass of NaNO₂ = 23 + 14 + 2$\times$16 = 69 g

2 NaNO₃ (s)$\to$ 2NaNO₂ (s) + O₂ (g)
2$\times$85=170g 2$\times$69=138g

170 g Pb(NO₃)₂ give residue = 138 g
Therefore x g Pb(NO₃)₂ will give residue = $\frac{138}{170}\times (5-\mathrm{x})\mathrm{g}$
= 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 = $\frac{72}{100}\times 5$
= 3.6 g
$\therefore$ 0.674x + 0.812 (5-x)= 3.6
0.138 = 0.46
x = $\frac{0.46}{0.138}$
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