Question

A solid mixture (5 gm) consists of lead nitrate and sodium nitrate, was heated below ${600}^{\circ }C$ until weight of residue was constant. If the loss in weight is 28% find amount of lead nitrate and sodium nitrate in the mixture.

• A. 3.32gm, 1.68gm
• B. 1.68gm, 3.32gm
• C. 4.2gm,1.6gm
• D. 6.6gm, 0.8gm

Answer 1

The correct option is A 3.32gm, 1.68gm

A solid mixture (5 gm) consists of lead nitrate and sodium nitrate
Let x g of this is lead nitrate. $5-x$ g of this will be sodium nitrate.
The molar mass of lead nitrate is 331.2 g/mol.
The molar mass of sodium nitrate is 85 g/mol.
The number of moles of lead nitrate $=\frac{\text{(x) g}}{\text{331.2 g/mol}}=\frac{\text{(x)}}{\text{331.2}}\text{mol}$.
The number of moles of sodium nitrate $=\frac{\text{(5-x) g}}{\text{85 g/mol}}=\frac{\text{(5-x)}}{\text{85}}\text{mol}$.
The thermal decomposition of lead nitrate is represented by the following equation.
$2Pb\left(N{O}_3{)}_2\right(s)\to 2PbO(s)+4N{O}_2(g)+O_2(g)$
Loss in weight for the thermal decomposition of lead nitrate is $\frac{\text{(x)}}{\text{331.2}}\text{mol}\times \text{( molar mass of}Pb(N{O}_3{)}_2-\text{molar mass of PbO)}$
Loss in weight for the thermal decomposition of lead nitrate is $\frac{\text{(x)}}{\text{331.2}}\text{mol}\times \text{( 331.2 g/mol}-\text{223.2 g/mol)}$
Loss in weight for the thermal decomposition of lead nitrate is $\frac{\text{(x)}}{\text{331.2}}\text{mol}\times \text{( molar mass of 108 g/mol )}$
Loss in weight for the thermal decomposition of lead nitrate is $\text{(0.326x) g}$
The thermal decomposition of sodium nitrate is represented by the following equation.
$2NaN{O}_3\to O_2+2NaN{O}_2$
Loss in weight for the thermal decomposition of sodium nitrate is $\frac{\text{(5-x)}}{\text{85}}\text{mol}\times \text{16 g/mol}$
Loss in weight for the thermal decomposition of sodium nitrate is $\text{0.188(5-x)}$
Total loss in weight for the thermal decomposition of lead nitrate and sodium nitrate is
$\text{(0.326x) g}+\text{0.188(5-x)}$
$=0.326x+0.9412-0.188x$
$=0.138x+0.9412$ g.
Percent loss in weight for the thermal decomposition of lead nitrate and sodium nitrate is
$\frac{0.138x+0.9412}{5}\times 100=2.76x+18.8$ %
But it is equal to 28 %.
$2.76x+18.8=28$
$2.76x=9.176$
$x=3.32$ g
Hence, the amount of lead nitrate in the mixture is 3.32 g.

The amount of sodium nitrate in the mixture is $5-3.32=1.68$.