Đáp án:
D
Giải thích các bước giải:
\(\begin{array}{l}
A{l_2}{(S{O_4})_3} + 6NaOH \to 2Al{(OH)_3} + 3N{a_2}S{O_4}\\
NaOH + Al{(OH)_3} \to NaAl{O_2} + 2{H_2}O\\
2NaAl{O_2} + C{O_2} + 3{H_2}O \to 2Al{(OH)_3} + N{a_2}C{O_3}\\
2Al{(OH)_3} \to A{l_2}{O_3} + 3{H_2}O\\
nA{l_2}{O_3} = \dfrac{{2,04}}{{102}} = 0,02\,mol\\
BTNT\,Al:nA{l_2}{(S{O_4})_3} = nA{l_2}{O_3} = 0,02\,mol\\
{C_M}A{l_2}{(S{O_4})_3} = \dfrac{{0,02}}{{0,02}} = 1M
\end{array}\)
