Đáp án:
\(\begin{array}{l}
a)\\
\% {m_{Al}} = 11,69\% \\
\% {m_{A{l_2}{O_3}}} = 88,31\% \\
b)\\
{V_{{\rm{dd}}HN{O_3}}} = 0,8l\\
c)\\
{C_M}Al{(N{O_3})_3} = 0,625M
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
Al + 4HN{O_3} \to Al{(N{O_3})_3} + NO + 2{H_2}O\\
A{l_2}{O_3} + 6HN{O_3} \to 2Al{(N{O_3})_3} + 3{H_2}O\\
{n_{NO}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{Al}} = {n_{NO}} = 0,1\,mol\\
{m_{Al}} = 0,1 \times 27 = 2,7g\\
{m_{A{l_2}{O_3}}} = 23,1 - 2,7 = 20,4g\\
\% {m_{Al}} = \dfrac{{2,7}}{{23,1}} \times 100\% = 11,69\% \\
\% {m_{A{l_2}{O_3}}} = 100 - 11,69 = 88,31\% \\
b)\\
{n_{A{l_2}{O_3}}} = \dfrac{{20,4}}{{102}} = 0,2\,mol\\
{n_{HN{O_3}}} = 0,1 \times 4 + 0,2 \times 6 = 1,6\,mol\\
{V_{{\rm{dd}}HN{O_3}}} = \dfrac{{1,6}}{2} = 0,8l\\
c)\\
{n_{Al{{(N{O_3})}_3}}} = 0,1 + 0,2 \times 2 = 0,5\,mol\\
{C_M}Al{(N{O_3})_3} = \dfrac{{0,5}}{{0,8}} = 0,625M
\end{array}\)
