Đáp án:
\(\begin{array}{l}
b)\\
{C_M}N{a_2}C{O_3} = 0,5M\\
c)\\
\% {m_{C{H_3}COOH}} = 60\% \\
\% {m_{{C_2}{H_5}OH}} = 40\% \\
d)\\
{m_{CaC{O_3}}} = 10g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2C{H_3}COOH + N{a_2}C{O_3} \to 2C{H_3}COONa + C{O_2} + {H_2}O\\
b)\\
{n_{C{O_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{N{a_2}C{O_3}}} = {n_{C{O_2}}} = 0,1\,mol\\
{C_M}N{a_2}C{O_3} = \dfrac{{0,1}}{{0,2}} = 0,5M\\
c)\\
{n_{C{H_3}COOH}} = 2{n_{C{O_2}}} = 0,2\,mol\\
\% {m_{C{H_3}COOH}} = \dfrac{{0,2 \times 60}}{{20}} \times 100\% = 60\% \\
\% {m_{{C_2}{H_5}OH}} = 100 - 60 = 40\% \\
d)\\
{n_{Ca{{(OH)}_2}}} = \dfrac{{74 \times 20\% }}{{74}} = 0,2\,mol\\
T = \frac{{0,2}}{{0,1}} = 2 \Rightarrow \text{ Tạo muối $CaCO_3$ } \\
Ca{(OH)_2} + C{O_2} \to CaC{O_3} + {H_2}O\\
{n_{CaC{O_3}}} = {n_{C{O_2}}} = 0,1\,mol\\
{m_{CaC{O_3}}} = 0,1 \times 100 = 10g
\end{array}\)
