nSO2=6,72/22,4=0,3mol
Gọi nFe=a, nAl=b

2Fe+6H2SO4→Fe2(SO4)3+3SO2+6H2O

a                                            1,5a

2Al+6H2SO4→Al2(SO4)3+3SO2+6H2O

b                                          1,5b

56a+27b=8,3

1,5a+1,5b=0,3

→a=0,1

   b=0,1

mFe=0,1.56=5,6g

%mFe=5,6/8,3.100%=67,47%

%mAl=100-67,47=32,53%

Đáp án:

\(\begin{array}{l}
\% Fe = 67,47\% \\
\% Al = 32,53\% 
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}
2Fe + 6{H_2}S{O_4} \to F{e_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
2Al + 6{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3S{O_2} + 6{H_2}O\\
{n_{S{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
hh:Fe(a\,mol);Al(b\,mol)\\
\left\{ \begin{array}{l}
56a + 27b = 8,3\\
\dfrac{3}{2}a + \dfrac{3}{2}b = 0,3
\end{array} \right.\\
 \Rightarrow a = 0,1;b = 0,1\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
\% Fe = \dfrac{{5,6}}{{8,3}} \times 100\%  = 67,47\% \\
\% Al = 100 - 67,47 = 32,53\% 
\end{array}\)