Đáp án:
\(\begin{array}{l}
a)\\
\% Al = 95,5\% \\
\% A{l_2}{O_3} = 4,5\% \\
b)\\
{V_{NO}} = 2,24l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
4Al + 3{O_2} \to 2A{l_2}{O_3}\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,1mol\\
{m_{Al}} = 0,1 \times 27 = 2,7g\\
{m_{A{l_2}{O_3}}} = 2,827 - 2,7 = 0,127g\\
\% Al = \dfrac{{2,7}}{{2,827}} \times 100\% = 95,5\% \\
\% A{l_2}{O_3} = 100 - 95,5 = 4,5\% \\
b)\\
Al + 4HN{O_3} \to Al{(N{O_3})_3} + NO + 2{H_2}O\\
A{l_2}{O_3} + 6HN{O_3} \to 2Al{(N{O_3})_3} + 3{H_2}O\\
{n_{NO}} = {n_{Al}} = 0,1mol\\
{V_{NO}} = 0,1 \times 22,4 = 2,24l
\end{array}\)