Đáp án:

a)10,375g

b)1,68l

c)12%

Giải thích các bước giải:

\(\begin{array}{l}
a)NaC{O_3} + 2C{H_3}{\rm{COOH}} \to 2C{H_3}{\rm{COO}}Na + C{O_2} + {H_2}O\\
{m_{C{H_3}COOH}} = \dfrac{{50 \times 30\% }}{{100\% }} = 15g \to {n_{C{H_3}COOH}} = 0,25mol\\
 \to {n_{NaC{O_3}}} = \dfrac{1}{2}{n_{C{H_3}COOH}} = 0,125mol\\
 \to {m_{NaC{O_3}}} = 0,125 \times 83 = 10,375g\\
b)Mg + 2C{H_3}{\rm{COOH}} \to Mg{(C{H_3}{\rm{COO)}}_2}{\rm{ + }}{{\rm{H}}_2}\\
{n_{Mg}} = 1mol\\
{m_{C{H_3}COOH}} = \dfrac{{30 \times 30\% }}{{100\% }} = 9g \to {n_{C{H_3}COOH}} = 0,15mol\\
 \to {n_{Mg}} > {n_{C{H_3}COOH}} \to {n_{{H_2}}} = \frac{1}{2}{n_{C{H_3}COOH}} = 0,075mol\\
 \to {V_{{H_2}}} = 0,075 \times 22,4 = 1,68l\\
c)Fe + 2C{H_3}{\rm{COOH}} \to Fe{(C{H_3}{\rm{COO)}}_2}{\rm{ + }}{{\rm{H}}_2}\\
{n_{{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
 \to {n_{C{H_3}COOH}} = 2{n_{{H_2}}} = 0,2mol\\
 \to {m_{C{H_3}COOH}} = 0,2 \times 60 = 12g\\
 \to C{\% _{{\rm{dd}}C{H_3}COOH}} = \dfrac{{12}}{{100}} \times 100\%  = 12\% 
\end{array}\)