Question

A solid mixture weighing $5$ g consisting of lead nitrate and sodium nitrate was heated below ${600}^o$C, until the mass of the residue was constant. If the loss in mass is $28\%$, the amount of sodium nitrate in the mixture is (as the nearest integer) :

Answer 1

$Pb{\left(N{O}_3\right)}_2\stackrel{\Delta }{\to }PbO+2N{O}_2\uparrow +\frac{1}{2}{O}_2\uparrow$
$a$ g
$NaN{O}_3\to NaN{O}_2+\frac{1}{2}{O}_2\uparrow$
$b$ g
a+b = 5$...\left(i\right)$
Given loss in mass of mixture is $28\%$ and thus, residue left after heating $=5-(5\times \frac{28}{100})=3.6g$
The residue left is $PbO+NaN{O}_3$
Moles of $Pb{N{O}_3}_2=$ Moles of $PbO$
$\frac{a}{331}=\frac{w_{PbO}}{223}$
$w_{PbO}=223\times \frac{a}{331}$
Similarly, Mole of $NaN{O}_3=$ Mole of $NaN{O}_2$
$\frac{b}{85}=\frac{w_{NaN{O}_2}}{69}$
$w_{NaN{O}_2}=69\times \frac{b}{85}$
$\frac{223\times a}{331}+\frac{69\times b}{85}=3.6...\left(ii\right)$
Therefore, by Eqs. (i) and (ii),
$a=3.32g;b=1.68g$
Sodium nitrate in mixture $=$ $2$ g.