Đáp án:
\(\begin{array}{l}
a.\\
{m_{Al}} = 10,8g\\
{m_{A{l_2}{O_3}}} = 10,2g\\
b.\\
{m_{NaAl{O_2}}} = 49,2g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2Al + 2{H_2}O + 2NaOH \to 2NaAl{O_2} + 3{H_2}\\
A{l_2}{O_3} + 2NaOH \to 2NaAl{O_2} + {H_2}O
\end{array}\)
\(\begin{array}{l}
a.\\
{n_{{H_2}}} = 0,6mol\\
\to {n_{Al}} = \dfrac{2}{3}{n_{{H_2}}} = 0,4mol\\
\to {m_{Al}} = 10,8g\\
\to {m_{A{l_2}{O_3}}} = 10,2g \to {n_{A{l_2}{O_3}}} = 0,1mol
\end{array}\)
\(\begin{array}{l}
b.\\
{n_{NaAl{O_2}}} = {n_{Al}} + 2{n_{A{l_2}{O_3}}} = 0,6mol\\
\to {m_{NaAl{O_2}}} = 49,2g
\end{array}\)