S Block Elements Question 4
Question 4 - 25 January - Shift 1
Compound A reacts with $NH_4 Cl$ and forms a compound B. Compound B reacts with $H_2 O$ and excess of $CO_2$ to form compound $C$ which on passing through or reaction with saturated $NaCl$ solution forms sodium hydrogen carbonate.
Compound A. B and C, are respectively.
(1) $CaCl_2, NH_3, NH_4 HCO_3$
(2) $CaCl_2, NH_4^{+},(NH_4)_2 CO_3$
(3) $Ca(OH)_2, NH_3, NH_4 HCO_3$
(4) $Ca(OH)_2, NH_4^{+},(NH_4)_2 CO_3$
Show Answer
Answer: (3)
Solution:
Formula: Reactions of s block elements
$ Ca(OH)_2+2 NH_4 Cl \xrightarrow{\Delta} 2 NH_3+CaCl_2+2 H_2 O $
$NH_3+H_2 O+CO_2 \longrightarrow NH_4 HCO_3$
(B)
$NH_4 HCO_3+NaCl \longrightarrow NaHCO_3 \downarrow+NH_4 Cl$
(C)
