Some Basic Concepts of Chemistry - Result Question 52
####51. A solid mixture $(5.0 g)$ consisting of lead nitrate and sodium nitrate was heated below $600^{\circ} C$ 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.
$(1990,4 M)$
Show Answer
Solution:
- Heating below $600^{\circ} C$ converts $Pb\left(NO _3\right) _2$ into $PbO$ but to $NaNO _3$ into $NaNO _2$ as
MW
$$ \begin{array}{cl} Pb\left(NO _3\right) _2 & \stackrel{\Delta}{\longrightarrow} PbO(s)+2 NO _2 \uparrow+\frac{1}{2} O _2 \uparrow \ 330 & 222 \ NaNO _3 & \stackrel{\Delta}{\longrightarrow} NaNO _2(s)+\frac{1}{2} O _2 \uparrow \end{array} $$
MW : $\quad 330$
MW : $\quad 85 \quad 28 \quad 69$
Weight loss $=5 \times \frac{28}{100}=1.4 g$
$\Rightarrow$ Weight of residue left $=5-1.4=3.6 g$
Now, let the original mixture contain $x g$ of $Pb\left(NO _3\right) _2$.
$\because \quad 330 g Pb\left(NO _3\right) _2$ gives $222 g$ PbO
$\therefore \quad x g Pb\left(NO _3\right) _2$ will give $\frac{222 x}{330} g PbO$
Similarly, $85 g NaNO _3$ gives $69 g NaNO{ } _2$
$\Rightarrow \quad(5-x) g NaNO _3$ will give $\frac{69(5-x)}{85} g NaNO _2$
$\Rightarrow$ Residue : $\frac{222 x}{330}+\frac{69(5-x)}{85}=3.6 g$
Solving for $x$ gives, $\quad x=3.3 g Pb\left(NO _3\right) _2$
$\Rightarrow \quad NaNO _3=1.7 g$.
