Đáp án:

 a) \({m_{C{H_3}COONa}} = 21,867{\text{ gam}}\)

b) \({{\text{V}}_{KOH}} = 2{\text{ lít}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(C{H_3}COONa + NaOH\xrightarrow{{CaO,{t^o}}}C{H_4} + N{a_2}C{O_3}\)

\({n_{C{H_4}}} = \frac{{4,48}}{{22,4}} = 0,2{\text{ mol = }}{{\text{n}}_{C{H_3}{\text{COONa lý thuyết}}}} \to {n_{C{H_3}COONa{\text{ thực tế}}}} = \frac{{0,2}}{{75\% }} = \frac{4}{{15}}{\text{ mol}}\)

\(\to {m_{C{H_3}COONa}} = \frac{4}{{15}}.(60 + 22) = 21,867{\text{ gam}}\)

\(C{H_4} + C{l_2}\xrightarrow{{}}C{H_3}Cl + HCl\)

\( \to {n_{HCl}} = {n_{C{H_4}}} = 0,2{\text{ mol}}\)

\(HCl + KOH\xrightarrow{{}}KCl + {H_2}O\)

\( \to {n_{KOH}} = {n_{HCl}} = 0,2{\text{ mol}} \to {{\text{V}}_{KOH}} = \frac{{0,2}}{{0,1}} = 2{\text{ lít}}\)