Sodium carbonate is generally prepared by Solvay process. In this process, $$N{H_3}$$  is recovered when the solution containing $$N{H_4}Cl$$   is treated with $$Ca{\left( {OH} \right)_2}.$$   Calcium chloride is obtained as a by - product.
$$2N{H_4}Cl + Ca{\left( {OH} \right)_2} \to $$     $$2N{H_3} + CaC{l_2} + 2{H_2}O$$

As the size of the alkali metal cation increases, thermal stability of their hydrides decreases.
Hence, the correct order of thermal stability of alkali metal hydrides is $$LiH > NaH > KH > RbH > CsH$$

No explanation is given for this question. Let's discuss the answer together.

Order of thermal stability is $${K_2}C{O_3} > N{a_2}C{O_3} > CaC{O_3} > MgC{O_3}$$
Hence, $$MgC{O_3}$$  releases $$C{O_2}$$  most easily \[MgC{{O}_{3}}\xrightarrow{\Delta }MgO+C{{O}_{2}}\]

Due to much lower freezing point of eutectic mixture of $$\frac{{CaC{l_2}}}{{{H_2}O}}.$$

Higher the lattice enthalpy lower will be solubility i.e.,
$${\text{lattice enthalpy}} \propto \frac{1}{{{\text{Solubility}}}}$$
Since the lattice enthalpy of alkali metals follow the order
$$Li > Na > K > Rb$$
Hence the correct order of solubility is
$$LiF < NaF < KF < RbF$$

The basic character of oxides increases down the group.

$$\eqalign{ & CaO + C{O_2} \to CaC{O_3} \cr & CaO + {H_2}O \to Ca{\left( {OH} \right)_2} \cr & {\text{hissing sound and}}\,\Delta H = - ve \cr} $$

All the elements except beryllium combine with hydrogen upon heating to form their hydrides, $$M{H_2}.$$
$$Be{H_2},$$  however, can be prepared by the reaction of $$BeC{l_2}$$  with $$LiAl{H_4}.$$
$$2BeC{l_2} + LiAl{H_4} \to $$     $$2Be{H_2} + LiCl + AlC{l_3}$$