Question

A solid mixture of weight $5$ g consisting of lead nitrate and sodium nitrate was heated below ${600}^{o}C$ until the weight of the residue was constant. If the loss in weight is $28.0\%$, the amount of lead nitrate in the mixture is :

• A. $3.32$ g
• B. $1.68$ g
• C. $3.87$ g
• D. $4.12$ g

Answer 1

Answer: (A)

$3.32$ g

Let the mass of $Pb(N{O}_3{)}_2$ be $a$ g.

The reaction is given below:
$Pb(N{O}_3{)}_2\to PbO+2N{O}_2\uparrow +\frac{1}{2}{O}_2\uparrow$
For $331$ g $Pb(N{O}_3{)}_2$, loss in weight $=108$ g
For $a$ g $Pb(N{O}_3{)}_2$, loss in weight $=\frac{108a}{331}$ g
Let the mass of $NaN{O}_3$ be $b$ g.
$NaN{O}_3\to NaN{O}_2,+\frac{1}{2}{O}_2\uparrow$
For $331$ g $NaN{O}_3$, loss in weight $=16$ g
For $b$ g $NaN{O}_3$, loss in weight $=\frac{16b}{85}$ g
Loss in weight$=\frac{5}{100}\times 28$ g (Given)
$\therefore \frac{108a}{331}+\frac{16b}{85}=\frac{5}{100}\times 28$ ....(i)

Also, $a+b=5$ ....(ii)
On solving equations (i) and (ii), we get $a=3.32$ g and $b=1.68$ g.