Đáp án:

\(\begin{array}{l}
\% Al = 20,9\% \\
\% A{l_2}{O_3} = 79,1\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1)\\
4Al + 3{O_2} \to 2A{l_2}{O_3}(1)\\
2)\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
{n_{A{l_2}{O_3}}} = \dfrac{m}{M} = \dfrac{{20,4}}{{102}} = 0,2mol\\
{n_{A{l_2}{O_3}(1)}} = \dfrac{2}{3}{n_{{O_2}}} = 0,1mol\\
{n_{A{l_2}{O_3}(2)}} = {n_{A{l_2}{O_3}}} - {n_{A{l_2}{O_3}(1)}} = 0,1mol\\
{m_{A{l_2}{O_3}(2)}} = n \times M = 0,1 \times 102 = 10,2g\\
{n_{Al}} = \dfrac{2}{3}{n_{{O_2}}} = 0,1mol\\
{m_{Al}} = n \times M = 0,1 \times 27 = 2,7g\\
\% Al = \dfrac{{2,7}}{{2,7 + 10,2}} \times 100\%  = 20,9\% \\
\% A{l_2}{O_3} = 100 - 20,9 = 79,1\% 
\end{array}\)

$\text{PTHH:}$

$4Al+3O_2\xrightarrow{t^0}2Al_2O_3$

     0,2       __   0,15                    (mol)

$m_{Al}=0.2.27=5,4(g)$

$\to$ Rắn duy nhất là $Al_2O_{3\text{..không pứ}}$

$\to n_{Al_2O_{\text{..kpứ}}}=\dfrac{20,4}{102}=0,2(mol)$

b) $m_{\text{..hhbđ}}=m_{Al}+m_{Al_2O_3}=5,4+20,4=25,8g$

$\to$ %$m_{Al}=\dfrac{\text{5,4.100%}}{25,8}\approx21$%

$\to$ %$m_{Al_2O_3}=\dfrac{\text{20,4.100%}}{25,8}=79$%