\(\begin{array}{l}
2)\\
a)\\
2KCl{O_3} \to 2KCl + 3{O_2}\\
C + {O_2} \to C{O_2}\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
CaC{l_2} + 2AgN{O_3} \to 2AgCl + Ca{(N{O_3})_2}\\
b)\\
FeS + 2HCl \to FeC{l_2} + {H_2}S\\
2{H_2}S + S{O_2} \to 3S + 2{H_2}O\\
S + 2Na \to N{a_2}S\\
N{a_2}S + ZnC{l_2} \to ZnS + 2NaCl\\
2{H_2}S + 3{O_2} \to 2S{O_2} + 2{H_2}O\\
2S{O_2} + {O_2} \to 2S{O_3}\\
S{O_3} + {H_2}O \to {H_2}S{O_4}\\
3)\\
Mg + {H_2}S{O_4} \to MgS{O_4} + {H_2}\\
Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\
n{H_2} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
hh:Mg(a\,mol),Fe(b\,mol)\\
24a + 56b = 6,8\\
a + b = 0,15\\
\Rightarrow a = 0,05;b = 0,1\\
\% mMg = \dfrac{{0,05 \times 24}}{{6,8}} \times 100\% = 17,65\% \\
\% mFe = 100 - 17,65 = 82,35\%
\end{array}\)
