Đáp án:

 20 C 

23 A

Giải thích các bước giải:

\(\begin{array}{l}
20)\\
mCu = 2(g) \Rightarrow mhh(Al,Fe) = 10 - 2 = 8,3g\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
n{H_2} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
hh:Al(a\,mol),Fe(b\,mol)\\
27a + 56b = 8,3\\
1,5a + b = 0,25\\
 \Rightarrow a = b = 0,1\,mol\\
\% mAl = \dfrac{{0,1 \times 27}}{{10,3}} \times 100\%  = 26,21\% \\
\% mCu = \dfrac{2}{{10,3}} \times 100\%  = 19,42\% \\
\% mFe = 100 - 26,21 - 19,42 = 54,37\% \\
23)\\
nFe = \dfrac{{2,8}}{{56}} = 0,05\,mol\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
nFeC{l_3} = nFeC{l_2} = nFe = 0,05\,mol\\
m = 0,05 \times 162,5 + 0,05 \times 127 = 14,475g
\end{array}\)