Chemistry - Redox Reactions Question with Solution | TestHub

Assume that the purity of KI is $x\%$.

Then, the pure mass of $\mathrm{KI}=\frac{10x}{100}=0.1xgm$.

The moles of $\mathrm{KI}=\frac{0.1x}{166}mol$.

 

The produced moles of ${\mathrm{I}}_2=\frac{0.1x}{166}\times \frac{4}{8}$.

We know that the moles of ${\mathrm{I}}_2=\left[Molesof\mathrm{NaI}\times \frac{1}{2}+Molesof{\mathrm{Cl}}_2\right]$.

$$\Rightarrow \frac{0.1x}{166}\times \frac{4}{8}=\left[\frac{0.3}{150}\times \frac{1}{2}+\frac{0.497}{71}\right]$$

$$\Rightarrow \frac{0.1x}{166\times 2}=0.001+0.007=0.008$$

$$x=26.56$$

The moles of ${\mathrm{I}}_2=0.008$.

 

The moles of ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3=MolesofNaI$.

$$\frac{20}{1000}\times M_{{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3}=0.002$$

$$M_{{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3}=0.1mol/L$$

We also know that $\frac{5}{8}\times [ReactedmolesofKIin1streaction]=[Givenmolesof{\mathrm{H}}_2{\mathrm{SO}}_4-Reactedmolesof{\mathrm{H}}_2{\mathrm{SO}}_{4}in{2}^{nd}reaction]$

 

. Assume the % (w/w) of ${\mathrm{H}}_2{\mathrm{SO}}_4$ is $w$.

$$\begin{array}{rl}\left[\frac{5}{8}\times 0.016\right]& =\left[10\times 1.47\times \frac{w}{100}\times \frac{1}{98}-\frac{1}{2}\times \frac{20}{100}\times 8\times \frac{1}{40}\right]\\ 0.01& =\left[\frac{0.147}{98}w-0.02\right]\\ \frac{0.147}{98}w& =0.03\\ w& =20\%\end{array}$$