(a)(i) Limestone is a cheap and available raw material for the solvay process. It is heated to decompose to produce carbon(IV) oxide, an important ingredient in the solvay reaction i.e CaCO\(_{3(S)}\) \(\to\) CaO\(_{(S)}\) + CO\(_{2(g)}\)
(ii) Ammoninia dissolved in water will give aqueous ammonia which constitute an important reactant in the solvay process.
(iii) Sodium chloride is the rich source of sodium which is one of the composing elements of NaHCO\(_3\), a main product of the solvay.process
i.e
(b) The hydrolysis reaction of sodium trioxocarbonate (IV) produces a basic medium
i.e 2Na\(^+\).CO\(^{2-}_3\) + 2H\(^+\)2OH\(^-\) -> 2NaOH + H\(_2\)CO\(^3\). Hence the reaction between dilute hydrochloric acid and sodium trioxocarbonate (IV) solution can be said to be acid-base reaction which is also neutralisation reaction.
(c) 2NaHCO\(_{3(S)}\) ---> Na\(_2\)CO\(_{3(s)}\) + H\(_{2}\)O\(_{(g)}\) + CO\(_{2(g)}\)
168g produced 106g Na\(_2\)CO\(_3\)
i.e lg will produce \(\frac{106}{168}\)
16.8g will produce \(\frac{106}{168} \times \frac{16.8}{1}\)
= 10.6g Na\(_2\)CO\(_3\)
