Ta có nH2 = \(\dfrac{7,84}{22,4}\) = 0,35 ( mol )
2Al + 3H2SO4 \(\rightarrow\) Al2(SO4)3 + 6H2
x............1,5x............x/2...............3x
Fe + H2SO4 \(\rightarrow\) FeSO4 + H2
y.........y.................y............y
=> \(\left\{{}\begin{matrix}27x+56y=13,9\\3x+y=0,35\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,04\\y=0,23\end{matrix}\right.\)
=> mAl = 27 . 0,04 = 1,08
=> %mAl = \(\dfrac{1,08}{13,9}\times100\approx\) 7,8 %
=> %mFe = 100 - 7,8 = 92,2%
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