Em tham khảo nha :
\(\begin{array}{l}
1)\\
a)\\
S{O_2} + {H_2}O \to {H_2}S{O_3}\\
S{O_2} + 2KOH \to {K_2}S{O_3} + {H_2}O\\
b)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
NaOH + HCl \to NaCl + {H_2}O\\
2)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = \dfrac{6}{{24}} = 0,25mol\\
{n_{HCl}} = 2{n_{Mg}} = 0,5mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,5}}{{0,25}} = 2M\\
{n_{{H_2}}} = {n_{Mg}} = 0,25mol\\
{V_{{H_2}}} = 0,25 \times 22,4 = 5,6l\\
3)\\
a)\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
b)\\
{n_{HCl}} = \dfrac{{200 \times 7,3\% }}{{36,5}} = 0,4mol\\
{n_{CaC{O_3}}} = \dfrac{{{n_{HCl}}}}{2} = 0,2mol\\
{m_{CaC{O_3}}} = 0,2 \times 100 = 20g\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,2mol\\
{V_{C{O_2}}} = 0,2 \times 22,4 = 4,48l\\
c)\\
{n_{CaC{l_2}}} = {n_{CaC{O_3}}} = 0,2mol\\
{m_{CaC{l_2}}} = 0,2 \times 111 = 22,2g\\
C{\% _{CaC{l_2}}} = \dfrac{{22,2}}{{20 + 200 - 0,2 \times 4}} \times 100\% = 10,5\%
\end{array}\)
