Đáp án:
\(\begin{array}{l}
b)\\
\% Al = 34,6\% \\
\% A{l_2}{O_3} = 65,4\% \\
c)\\
C{\% _{HCl}} = 20\% \\
{m_{AlC{l_3}}} = 5,34g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
b)\\
{n_{{H_2}}} = \dfrac{{0,672}}{{22,4}} = 0,03mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,02mol\\
{m_{Al}} = 0,02 \times 27 = 0,54g\\
{m_{A{l_2}{O_3}}} = 1,56 - 0,54 = 1,02g\\
\% Al = \dfrac{{0,54}}{{1,56}} \times 100\% = 34,6\% \\
\% A{l_2}{O_3} = 100 - 34,6 = 65,4\% \\
c)\\
{n_{A{l_2}{O_3}}} = \dfrac{{1,02}}{{102}} = 0,01mol\\
{n_{HCl}} = 3{n_{Al}} + 6{n_{A{l_2}{O_3}}} = 0,12mol\\
{m_{HCl}} = 0,12 \times 36,5 = 4,38g\\
C{\% _{HCl}} = \dfrac{{4,38}}{{21,9}} \times 100\% = 20\% \\
{n_{AlC{l_3}}} = {n_{Al}} + 2{n_{A{l_2}{O_3}}} = 0,04mol\\
{m_{AlC{l_3}}} = 0,04 \times 133,5 = 5,34g
\end{array}\)