JEE Main 26 Feb 2021 Shift 2 Solved Paper

(A) Sodium carbonate is manufactured by Solvay process $\Rightarrow$ (iv) of List-ll. Solvay process's reactions are as follows In Solvay's process, first ammonia $(\mathrm{NH}_3)$ reacts with $\mathrm{CO}_2$ and $\mathrm{H}_2\mathrm{O}$ to give ammonium bicarbonate $(\mathrm{NH}_4\mathrm{HCO}_3)$. This bicarbonate react with $\mathrm{NaCl}$ to give sodium bicarbonate $(\mathrm{NaHCO}_3)$ along with ammonium chloride $(\mathrm{NH}_4\mathrm{Cl})$. At last, $\mathrm{NaHCO}_3$ is heated to form sodium carbonate as major product. $\mathrm{NH}_3 + \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{NH}_4\mathrm{HCO}_3$ $\mathrm{NH}_4\mathrm{HCO}_3 + \underset{(\text{Brine})}{\mathrm{NaCl}} \longrightarrow \mathrm{NaHCO}_3 \downarrow + \mathrm{NH}_4\mathrm{Cl}$ $2\mathrm{NaHCO}_3 \rightarrow \Delta_2 \rightarrow \underset{\text{Sodiumcarbonate}}{\mathrm{Na}_2\mathrm{CO}_3} + \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}$ (B) Titanium is purified by van-Arkel method $\Rightarrow$ (iii) of List-II van-Arkel method Crude titanium $\xrightarrow[\mathrm{I}_2]{780\,\text{K}}$ Til $_{4}(g) \xrightarrow{1800\,\text{K}} \underset{(\text{Pure})}{\mathrm{Ti}} + 2\mathrm{I}_2$ (C) Chlorine is manufactured by Deacon process $\Rightarrow$ (i) of List-II. Deacon process $4\mathrm{HCl}(g) + \underset{[\text{Air}]}{\mathrm{O}_2(g)} \xrightarrow[723\,\text{K}]{\mathrm{CuCl}_2 \; \text{ (catalyst) }\;} \underset{\text{Chlorinegas}}{2\mathrm{Cl}_2} + 2\mathrm{H}_2\mathrm{O}(g)$ (D) Sodium hydroxide is manufactured by Castner-Kellner cell $\Rightarrow$ (ii) of List-II. Castner-Kellner cell reaction Electrolyte used is brine, saturated $\mathrm{NaCl}$ solution. Cathode $(\mathrm{Hg})$ : $\mathrm{Na}^+ + \mathrm{e}^- \longrightarrow \mathrm{Na} \xrightarrow{\mathrm{Hg}} \mathrm{Na}/\mathrm{Hg}$ Anode (carbon) : $\mathrm{Cl}^- + \mathrm{e}^- \longrightarrow \frac{1}{2}\mathrm{Cl}_2 \uparrow$ $2\mathrm{Na}/\mathrm{Hg} + 2\mathrm{H}_2\mathrm{O} \longrightarrow \underset{\text{Sodiumhydroxide}}{2\mathrm{NaOH}} + 2\mathrm{Hg} + \mathrm{H}_2 \uparrow$