Đáp án:
\(\begin{array}{l}
a)\\
{V_{C{O_2}}} = 6,72l\\
b)\\
{C_\% }HCl = 2,35\% \\
{C_\% }NaCl = 11,3\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
{n_{N{a_2}C{O_3}}} = \dfrac{{31,8}}{{106}} = 0,3\,mol\\
{n_{HCl}} = \dfrac{{292 \times 10\% }}{{36,5}} = 0,8\,mol\\
{n_{N{a_2}C{O_3}}} < \dfrac{{{n_{HCl}}}}{2} \Rightarrow HCl \text{ dư }\\
{n_{C{O_2}}} = {n_{N{a_2}C{O_3}}} = 0,3\,mol\\
{V_{C{O_2}}} = 0,3 \times 22,4 = 6,72l\\
b)\\
{m_{{\rm{dd}}spu}} = 31,8 + 292 - 0,3 \times 44 = 310,6g\\
{n_{HCl}} \text{ dư }= 0,8 - 0,3 \times 2 = 0,2\,mol\\
{n_{NaCl}} = 2{n_{N{a_2}C{O_3}}} = 0,6\,mol\\
{C_\% }HCl \text{ dư }= \dfrac{{0,2 \times 36,5}}{{310,6}} \times 100\% = 2,35\% \\
{C_\% }NaCl = \dfrac{{0,6 \times 58,5}}{{310,6}} \times 100\% = 11,3\%
\end{array}\)
