Em tham khảo nha :
\(\begin{array}{l}
a)\\
CaC{O_3} + 2C{H_3}COOH \to {(C{H_3}COO)_2}Ca + C{O_2} + {H_2}O\\
b)\\
{n_{CaC{O_3}}} = \dfrac{{20}}{{100}} = 0,2mol\\
{n_{C{H_3}COOH}} = 2{n_{CaC{O_3}}} = 0,4mol\\
{C_{{M_{C{H_3}COOH}}}} = \dfrac{{0,4}}{{0,4}} = 1M\\
c)\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
{n_{C{H_3}COOH}} = 0,2 \times 1 = 0,2mol\\
{n_{C{H_3}COO{C_2}{H_5}}} = {n_{C{H_3}COOH}} = 0,2mol\\
{m_{C{H_3}COO{C_2}{H_5}}} = 0,2 \times 88 = 17,6g\\
\text{Khối lượng este thu được là :}\\
m = \dfrac{{17,6 \times 80}}{{100}} = 14,08g
\end{array}\)
