Em tham khảo nha :
\(\begin{array}{l}
a)\\
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
{n_{N{a_2}C{O_3}}} = \dfrac{{200 \times 10,6\% }}{{106}} = 0,2mol\\
{n_{HCl}} = 2{n_{N{a_2}C{O_3}}} = 0,4mol\\
{m_{HCl}} = 0,4 \times 36,5 = 14,6g\\
C{\% _{HCl}} = \dfrac{{14,6}}{{100}} \times 100\% = 14,6\% \\
b)\\
{n_{C{O_2}}} = {n_{N{a_2}C{O_3}}} = 0,2mol\\
{V_{C{O_2}}} = 0,2 \times 22,4 = 4,48l\\
c)\\
{n_{NaCl}} = {n_{HCl}} = 0,4mol\\
{m_{NaCl}} = 0,4 \times 58,5 = 23,4g\\
C{\% _{NaCl}} = \dfrac{{23,4}}{{200 + 100 - 0,2 \times 44}} \times 100\% = 8,04\%
\end{array}\)
