Em tham khảo nha:

\(\begin{array}{l}
a)\\
2C{H_3}COOH + CaC{O_3} \to {(C{H_3}COO)_2}Ca + C{O_2} + {H_2}O\\
{n_{CaC{O_3}}} = \dfrac{{30}}{{100}} = 0,3\,mol\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,3\,mol\\
{V_{C{O_2}}} = 0,3 \times 22,4 = 6,72l\\
b)\\
{n_{C{H_3}COOH}} = 2{n_{CaC{O_3}}} = 0,6\,mol\\
{C_M}C{H_3}COOH = \dfrac{{0,6}}{{0,2}} = 3M\\
c)\\
{C_2}{H_5}OH + {O_2} \xrightarrow{\text{ men giấm }} C{H_3}COOH + {H_2}O\\
{n_{{C_2}{H_5}OH}} = {n_{C{H_3}COOH}} = 0,6\,mol\\
H = 65\%  \Rightarrow {m_{{C_2}{H_5}OH}} = \dfrac{{0,6 \times 46}}{{65\% }} = \dfrac{{552}}{{13}}(g)\\
{V_{{C_2}{H_5}OH}} = \dfrac{{\frac{{552}}{{13}}}}{{0,8}} = \dfrac{{690}}{{13}}\,ml\\
{V_{{\rm{dd}}{C_2}{H_5}OH}} = \dfrac{{\dfrac{{690}}{{13}}}}{{25}} \times 100 = 212,3\,ml
\end{array}\)