\(\begin{array}{l}
1)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O\\
nhh = \dfrac{{7,28}}{{22,4}} = 0,325\,mol\\
hh:Al(a\,mol),CaC{O_3}(b\,mol)\\
1,5a + b = 0,325\\
27a + 100b = 14,05\\
 \Rightarrow a = 0,15;b = 0,1\\
\% mAl = \dfrac{{0,15 \times 27}}{{14,05}} \times 100\%  = 28,83\% \\
\% mCaC{O_3} = 100 - 28,83 = 71,17\% \\
3)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
N{a_2}C{O_3} + 2HCl \to 2NaCl + C{O_2} + {H_2}O\\
nhh = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
hh:Fe(a\,mol),N{a_2}C{O_3}(b\,mol)\\
56a + 106b = 21,8\\
a + b = 0,3\\
 \Rightarrow a = 0,2;b = 0,1\\
\% mFe = \dfrac{{0,2 \times 56}}{{21,8}} \times 100\%  = 51,38\% \\
\% mN{a_2}C{O_3} = 100 - 51,38 = 48,62\% 
\end{array}\)