Find A, B and C in the following reactions :
\(\mathrm{NH_3}+A+\mathrm{CO_2}\rightarrow(\mathrm{NH_4})_2\mathrm{CO_3}\)
\((\mathrm{NH_4})_2\mathrm{CO_3}+\mathrm{H_2O}+B\rightarrow\mathrm{NH_4HCO_3}\)
\(\mathrm{NH_4HCO_3}+\mathrm{NaCl}\rightarrow\mathrm{NH_4Cl}+C\)
\(\mathrm{NH_3}+A+\mathrm{CO_2}\rightarrow(\mathrm{NH_4})_2\mathrm{CO_3}\)
\((\mathrm{NH_4})_2\mathrm{CO_3}+\mathrm{H_2O}+B\rightarrow\mathrm{NH_4HCO_3}\)
\(\mathrm{NH_4HCO_3}+\mathrm{NaCl}\rightarrow\mathrm{NH_4Cl}+C\)
Show Hint
- \(A-\mathrm{O_2};\ B-\mathrm{CO_2};\ C-\mathrm{Na_2CO_3}\)
- \(A-\mathrm{H_2O};\ B-\mathrm{O_2};\ C-\mathrm{NaHCO_3}\)
- \(A-\mathrm{H_2O};\ B-\mathrm{O_2};\ C-\mathrm{Na_2CO_3}\)
- \(A-\mathrm{H_2O};\ B-\mathrm{CO_2};\ C-\mathrm{NaHCO_3}\)
The Correct Option is D
Solution and Explanation
Step 1: \(2\mathrm{NH_3}+\mathrm{H_2O}+\mathrm{CO_2}\rightarrow(\mathrm{NH_4})_2\mathrm{CO_3}\), so A = H2O (atoms balance: 2 N, 8 H, 1 C, 3 O).
Step 2: \((\mathrm{NH_4})_2\mathrm{CO_3}+\mathrm{H_2O}+\mathrm{CO_2}\rightarrow2\mathrm{NH_4HCO_3}\), so B = CO2.
Step 3: \(\mathrm{NH_4HCO_3}+\mathrm{NaCl}\rightarrow\mathrm{NH_4Cl}+\mathrm{NaHCO_3}\), so C = NaHCO3 (the less soluble sodium hydrogencarbonate precipitates).
Hence A = H2O, B = CO2, C = NaHCO3.
Answer: (4) \(A-\mathrm{H_2O};\ B-\mathrm{CO_2};\ C-\mathrm{NaHCO_3}\)
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