Chemistry - Mole Concept Question with Solution | TestHub

$1\mathrm{M}\to 0.4\mathrm{M}(6\mathrm{mol}{\mathrm{HNO}}_3$ consumed $)$

$\mathrm{Pb}+4{\mathrm{HNO}}_3\to \mathrm{Pb}({\mathrm{NO}}_3{)}_2+2{\mathrm{NO}}_2+2{\mathrm{H}}_2\mathrm{O}$

$\mathrm{x}/4$ x x/2

$3\mathrm{Pb}+8{\mathrm{HNO}}_3\to 3\mathrm{Pb}({\mathrm{NO}}_3{)}_2+2\mathrm{NO}+4{\mathrm{H}}_2\mathrm{O}$

$\frac{3(6-x)}{8}6-x\frac{6-x}{4}$

$\frac{\frac{x}{4}}{\frac{3(6-x)}{8}}=4\Rightarrow x=\frac{36}{7}$

Hence, moles of Pb consumed $=\frac{3(6-x)}{8}+\frac{x}{4}=\frac{45}{28}\mathrm{mol}$.

Moles of ${\mathrm{NO}}_2=\frac{36}{7\times 2}=\frac{18}{7}\mathrm{mol}$.

Moles of ${\mathrm{NO}}_2$ escaping from solution $=\frac{18}{7}\times \frac{90}{100}=\frac{81}{35}$.

$\tan \theta =15=\frac{\mathrm{y}}{0.2}$

$\mathrm{y}=3$

Hence, the mole ratio of Pb

consumed between $0.2\mathrm{M}$ and the endpoint $=1$.

$\mathrm{Pb}+4{\mathrm{HNO}}_3\to \mathrm{Pb}({\mathrm{NO}}_3{)}_2+2{\mathrm{NO}}_2+2{\mathrm{H}}_2\mathrm{O}$

$\frac{\mathrm{x}}{4}$ x x/4

$3\mathrm{Pb}+8{\mathrm{HNO}}_3\stackrel{\mathrm{x}}{\to }3\mathrm{Pb}({\mathrm{NO}}_3{)}_2+2\mathrm{NO}+4{\mathrm{H}}_2\mathrm{O}$

$\frac{3(2-x)}{8}2-x\frac{3(2-x)}{8}$

$\frac{\frac{x}{4}}{\frac{3(2-x)}{8}}=1$

$2\mathrm{x}=6-3\mathrm{x}$

$5\mathrm{x}=6$

$\mathrm{x}=\frac{6}{5}$

Moles of $\mathrm{Pb}({\mathrm{NO}}_3{)}_2$ produced between point A and B $=\frac{\mathrm{X}}{4}+\frac{3(2-\mathrm{X})}{8}=0.6\mathrm{mol}$.