Đáp án:
a) 27,42%
b) 10%
c) 66,67%
Giải thích các bước giải:
\(\begin{array}{l}
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
a)\\
n{H_2} = \frac{{3,36}}{{22,4}} = 0,15\,mol\\
= > nAl = 0,1mol\\
= > mAl = 0,1 \times 27 = 2,7g\\
mA{l_2}{O_3} = 3,72 - 2,7 = 1,02g\\
\% mA{l_2}{O_3} = \frac{{1,02}}{{3,72}} \times 100\% = 27,42\% \\
b)\\
nHCl = 0,1 \times 3 + 0,01 \times 6 = 0,36\,mol\\
C\% HCl = \frac{{0,36 \times 36,5}}{{131,4}} \times 100\% = 10\% \\
c)\\
CuO + {H_2} \to Cu + {H_2}O\\
nCuOpu = amol\\
24 - 80a + 64a = 22,4\\
= > a = 0,1\,mol\\
H = \frac{{0,1}}{{0,15}} \times 100\% = 66,67\%
\end{array}\)