A solid mixture (5 gm) consisting of lead nitrate and sodium nitrate, was heated below 600 $^{o}C$ until weight residue was constant. If the loss in weight is 28%, find the amount of lead nitrate and sodium nitrate in the mixture.
A. 3.32 gm, 1.68 gm
B. 1.68 gm, 3.32 gm
C. 4.2 gm, 1.6 gm
D. 6.6 gm, 0.8 gm

Hint: The decomposition of lead nitrate at 600 $^{o}C$ is as follows.
\[2Pb{{(N{{O}_{3}})}_{2}}(s)\xrightarrow{{{600}^{o}}C}2PbO(s)+4N{{O}_{2}}(g)+{{O}_{2}}(g)\]
 The decomposition of Sodium nitrate at 600 $^{o}C$ is as follows.
\[2NaN{{O}_{3}}\xrightarrow{{{600}^{o}}c}2NaN{{O}_{2}}+{{O}_{2}}\]

Complete step by step solution:
- In the question it is given that a solid mixture (5 gm) consisting of lead nitrate and sodium nitrate, was heated below 600 $^{o}C$ until weight residue was constant.
 - Assume x gm of lead nitrate and 5-x gm of sodium nitrate is present in the given mixture.
- The molecular weight of lead nitrate = 331.2 gm.
- The molecular weight of sodium nitrate = 85 gm.
- Therefore the number of moles of lead nitrate is$\dfrac{x}{331.2}$ .
- The number of moles of lead nitrate is $\dfrac{5-x}{85}$
- The decomposition of lead nitrate at 600$^{o}C$ is as follows.
\[2Pb{{(N{{O}_{3}})}_{2}}(s)\xrightarrow{{{600}^{o}}C}2PbO(s)+4N{{O}_{2}}(g)+{{O}_{2}}(g)\]
- From the above chemical equation the loss in weight of the thermal decomposition of lead nitrate is as follows.
\[\begin{align}
& =\dfrac{\text{x}}{\text{331}\text{.2}}\text{ }\!\!\times\!\!\text{ (molar mass of Pb(N}{{\text{O}}_{\text{3}}}{{\text{)}}_{\text{2}}}\text{- }\!\!~\!\!\text{ molar }\!\!~\!\!\text{ mass }\!\!~\!\!\text{ of }\!\!~\!\!\text{ PbO)} \\
& =\dfrac{\text{x}}{\text{331}\text{.2}}\text{ }\!\!\times\!\!\text{ (331}\text{.2- }\!\!~\!\!\text{ 223}\text{.2)} \\
& =\dfrac{\text{x}}{\text{331}\text{.2}}\text{ }\!\!\times\!\!\text{ 108} \\
& \text{=0}\text{.326x gm} \\
\end{align}\]
- The decomposition of Sodium nitrate at 600 $^{o}C$ is as follows.
\[2NaN{{O}_{3}}\xrightarrow{{{600}^{o}}c}2NaN{{O}_{2}}+{{O}_{2}}\]
- From the above chemical equation the loss in weight of the thermal decomposition of sodium nitrate is as follows.
\[\begin{align}
& =\dfrac{\text{5-x}}{\text{85}}\text{ }\!\!\times\!\!\text{ (molar mass of NaN}{{\text{O}}_{\text{3}}}\text{- }\!\!~\!\!\text{ molar }\!\!~\!\!\text{ mass }\!\!~\!\!\text{ of }\!\!~\!\!\text{ NaN}{{\text{O}}_{\text{2}}}\text{)} \\
& =\dfrac{\text{5-x}}{\text{85}}\text{ }\!\!\times\!\!\text{ (16)} \\
& \text{=0}\text{.188(5-x)} \\
\end{align}\]
- Total loss in weight of the mixture after thermal decomposition of the mixture (lead nitrate and sodium nitrate) is = (0.326x) +0.188(5-x) = (0.138x+0.9412) gm.
- Percentage of loss in weight for the thermal decomposition of the mixture is
\[\begin{align}
& \text{=}\dfrac{\text{0}\text{.138x+0}\text{.9412}}{\text{5}}\text{ }\!\!\times\!\!\text{ 100} \\
& \text{=(2}\text{.76x+18}\text{.8)% } \\
\end{align}\]
- In the question it is given that the percentage of loss in weight for the thermal decomposition of the mixture is 28 %.
- Then
$2.76x+18.8 = 28$
$2.76x = 9.176$
$x = 3.32 gm$.
- Therefore the amount of lead nitrate in the 5gm of the mixture is 3.32 gm.
- Therefore the amount of sodium nitrate in the 5 gm of the mixture = 5 – 3.32 = 1.68 gm.

So the answer is option A.

Note: The sodium nitrate and lead nitrate after thermal decomposition lose some weight due to the liberation of nitrogen dioxide and oxygen by lead nitrate and liberation of oxygen by sodium nitrate. Maximum compounds lose some weight after heating due to the loss of some products.