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Grade upto college level Physical Chemistry

A solid mixture (5.0 g) consisting of lead nitrate and sodium nitrate was heated below 600oC until the weight of the residue was constant. If the loss in weight is 28.0 per cent, find the amount of lead nitrate and sodium nitrate in the mixture.

Profile image of Amit Saxena
12 Years agoGrade upto college level
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1 Answer

Profile image of Navjyot Kalra
12 Years ago
Let the amount of NaNO3 in the mixture = x g
∴ The amount of Pb(NO3)2 in the mixture = (5 - x) g
Heating effect of sodium nitrate and lead nitrate
2NaNO3 \overset{\Delta }{\rightarrow}2NaNO2 + O2
2(23+14+48) = 170 g 2 * 16 = 32 g
2 Pb(NO3)2 \overset{\Delta }{\rightarrow}
2(207 +28 +96) = 662 g
2PbO2+ 4NO2 + O2
\underbrace{4(14+32)=184 g 2*16=32 g}
216 g
Now since, 170 g of NaNO3 gives = 32 g of O2
∴ x g of NaNO3 gives = 32/170 * x g of O2
Similarly, 662 g of Pb(NO­3)2 gives = 216 g of gases
(5 - x) g of Pb(NO3)2 gives = 216/662 * (5 - x) g of gases
(NO2 + O2)
Actual loss, on heating, is 28% of 5 g of mixture
= 5 * 28/100 = 1.4 g
∴ 32 x/170 + 216/662 * (5 -x) = 1.4
32x * 662 + 216(5 -x) * 170 = 1.4 * 170 * 662
21184 x + 183600 – 36720 x = 157556
- 15536 x = - 26044
X = 1.676 g
Wt. of NaNO3 = 1.676 g
And Wt. of Pb(NO3)2 = 5 – 1.676 g = 3.324 g

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