Đáp án:
\(\begin{array}{l}
a){m_{{\rm{dd}}N{a_2}C{O_3}}} = 126,19g\\
b){V_{C{O_2}}} = 2,24l\\
c)C{\% _{C{H_3}{\rm{COO}}Na}} = 7,39\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2C{H_3}{\rm{COO}}H + N{a_2}C{O_3} \to 2C{H_3}{\rm{COO}}Na + {H_2}O + C{O_2}\\
{m_{C{H_3}{\rm{COO}}H}} = \dfrac{{100 \times 12\% }}{{100\% }} = 12g\\
\to {n_{C{H_3}{\rm{COO}}H}} = 0,2mol
\end{array}\)
\(\begin{array}{l}
a)\\
{n_{N{a_2}C{O_3}}} = \dfrac{1}{2}{n_{C{H_3}{\rm{COO}}H}} = 0,1mol\\
\to {m_{N{a_2}C{O_3}}} = 10,6g\\
\to {m_{{\rm{dd}}N{a_2}C{O_3}}} = \dfrac{{10,6}}{{8,4\% }} \times 100\% = 126,19g
\end{array}\)
\(\begin{array}{l}
b)\\
{n_{C{O_2}}} = \dfrac{1}{2}{n_{C{H_3}{\rm{COO}}H}} = 0,1mol\\
\to {V_{C{O_2}}} = 2,24l
\end{array}\)
\(\begin{array}{l}
c)\\
{n_{C{H_3}{\rm{COO}}Na}} = {n_{C{H_3}{\rm{COO}}H}} = 0,2mol\\
\to {m_{C{H_3}{\rm{COO}}Na}} = 16,4g\\
{m_{{\rm{dd}}}} = {m_{{\rm{dd}}C{H_3}{\rm{COO}}H}} + {m_{{\rm{dd}}N{a_2}C{O_3}}} - {m_{C{O_2}}} = 221,79g\\
\to C{\% _{C{H_3}{\rm{COO}}Na}} = \dfrac{{16,4}}{{221,79}} \times 100\% = 7,39\%
\end{array}\)
