Đáp án:

\({m_{dd{\text{NaHC}}{{\text{O}}_3}}} = 168{\text{ gam}}\)

\(C{\% _{C{H_3}COONa}} = 4,84\% \)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(C{H_3}COOH + NaHC{O_3}\xrightarrow{{}}C{H_3}COONa + C{O_2} + {H_2}O\)

Ta có: 

\({m_{C{H_3}COOH}} = 200.6\%  = 12{\text{ gam}} \to {{\text{n}}_{C{H_3}COOH}} = {n_{NaHC{O_3}}} = \frac{{12}}{{60}} = 0,2{\text{ mol}} \to {{\text{m}}_{NaHC{O_3}}} = 0,2.(23 + 1 + 60) = 16,8{\text{ gam}}\)

\( \to {m_{dd{\text{NaHC}}{{\text{O}}_3}}} = \frac{{16,8}}{{10\% }} = 168{\text{ gam}}\)

Theo phản ứng: 

\({n_{C{H_3}COONa}} = {n_{C{O_2}}} = {n_{C{H_3}COOH}} = 0,2{\text{ mol}} \to {{\text{m}}_{C{H_3}COONa}} = 0,2.(60 + 22) = 17,6{\text{ gam}}\)

BTKL: 

\({m_{dd{\text{ sau phản ứng}}}} = {m_{dd\;C{H_3}COOH}} + {m_{dd{\text{ NaHC}}{{\text{O}}_3}}} -{m_{C{O_2}}}= 200 + 168 - 0,2.44 = 359,2{\text{ gam}}\)

\( \to C{\% _{C{H_3}COONa}} = \frac{{17,6}}{{359,2}} = 4,84\% \)

\(\begin{array}{l}
a)\\
n_{CH_3COOH}=\frac{200.6\%}{60}=0,2(mol)\\
CH_3COOH+NaHCO_3\to CH_3COONa+CO_2+H_2O\\
n_{NaHCO_3}=n_{CH_3COOH}=0,2(mol)\\
m_{dd\,NaHCO_3}=\frac{0,2.84}{10\%}=168(g)\\
b)\\
n_{CH_3COONa}=n_{CO_2}=0,2(mol)\\
C\%_{CH_3COONa}=\frac{0,2.82}{200+168-0,2.44}.100\%\approx 4,9\%
\end{array}\)