Đáp án:
a) 54% 24% và 22%
c) 1,6l
Giải thích các bước giải:
\(\begin{array}{l}
n{H_2} = \dfrac{{8,96}}{{22,4}} = 0,4\,mol\\
nCuO = \dfrac{{2,75}}{{80}} = 0,034375\,mol\\
hh:Al(a\,mol),Mg(b\,mol),Cu(c\,mol)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
a\,\,\,\,\,\,\,\,\,\,\,\,\,3a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1,5a\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
b\,\,\,\,\,\,\,\,\,\,\,\,\,2b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\\
2Cu + {O_2} \to 2CuO\\
c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\\
27a + 24b + 64c = 10\\
1,5a + b = 0,4\\
c = 0,034375\\
= > a = 0,2;b = 0,1;c = 0,034375\\
\% mAl = \dfrac{{0,2 \times 27}}{{10}} \times 100\% = 54\% \\
\% mMg = \dfrac{{0,1 \times 24}}{{10}} \times 100\% = 24\% \\
\% mCu = 100 - 54 - 24 = 22\% \\
b)\\
nHCl = 3 \times 0,2 + 2 \times 0,1 = 0,8\,mol\\
VHCl = \dfrac{{0,8}}{{0,5}} = 1,6l
\end{array}\)
