Ta có nH2 = \(\dfrac{7,168}{22,4}\) = 0,32 ( mol )
2Al + 6HCl \(\rightarrow\) 2AlCl3 + 3H2
x...........3x..........x..............1,5x
Fe + 2HCl \(\rightarrow\) FeCl2 + H2
y.........2y..........y.............y
=> \(\left\{{}\begin{matrix}1,5+y=0,32\\27x+56y=8,8\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=0,16\\y=0,08\end{matrix}\right.\)
=> mAl = 27 . 0,16 = 4,32 ( gam )
=> %mAl = \(\dfrac{4,32}{8,8}\times100\approx49,09\%\)
=> %mFe = 100 - 49,09 = 50,91 %
=> mHCl = ( 3x + 2y ) . 36,5
=> mHCl = ( 3 . 0,16 + 2 . 0,08 ) . 36,5 = 23,36 ( gam )
2Al + 3H2SO4 \(\rightarrow\) Al2(SO4)3 + 3H2
0,16....0,24
Fe + H2SO4 \(\rightarrow\) FeSO4 + H2
0,08...0,08
=> mH2SO4 = ( 0,24 + 0,08 ) . 98 = 31,36 ( gam )
