Ta có nH2 = \(\dfrac{14,56}{22,4}\) = 0,65 ( mol )
2Al + 3H2SO4 \(\rightarrow\) Al2(SO4)3 + 3H2
x...........1,5x..............x/2...............1,5x
Fe + H2SO4 \(\rightarrow\) FeSO4 + H2
y........y....................y..........y
=> \(\left\{{}\begin{matrix}27x+56y=19,3\\1,5x+y=0,65\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,3\\y=0,2\end{matrix}\right.\)
=> mAl = 27 . 0,3 = 8,1 ( gam )
=> mFe = 19,3 - 8,1 = 11,2 ( gam )