Đáp án:
\(\begin{array}{l}
{n_{Al}} = 0,1\,mol\\
{n_{Mg}} = 0,2\,mol
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
A{l_2}{O_3} + 2NaOH \to 2NaAl{O_2} + {H_2}O\\
2Al + 2NaOH + 2{H_2}O \to 2NaAl{O_2} + 3{H_2}\\
{n_{Al}} = \dfrac{{0,15 \times 2}}{3} = 0,1\,mol\\
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}(1)\\
Mg + 2HCl \to MgC{l_2} + {H_2}(2)\\
A{l_2}{O_3} + 6HCl \to 2AlC{l_3} + 3{H_2}O(3)\\
{n_{{H_2}(2)}} = 0,35 - 0,15 = 0,2\,mol\\
{n_{Mg}} = {n_{{H_2}(2)}} = 0,2\,mol
\end{array}\)