2Al +3H2SO4---->Al2(SO4)3 +3H2(1)
Fe +H2SO4 ---->Fe2(SO4) +H2(2)
a) Gọi n\(_{Al}=x\Rightarrow m_{Al}=27x\)
n\(_{Fe}=y\Rightarrow m_{Fe}=56y\)
=> 27x+56y=11(*)
Mặt khác:
n\(_{H2}=\frac{8,96}{22,4}=0,1\left(mol\right)\)
Theo pthh1
n\(_{H2}=\)\(\frac{3}{2}n_{Al}=1,5x\left(mol\right)\)
Theo PTHH2
n\(_{H2}=n_{Fe}=y\left(mol\right)\)
=> 1,5x+y=0,4(**)
Từ (*)và(**) ta có hệ pt
\(\left\{{}\begin{matrix}27x+56y=11\\1,5x+y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
%m\(_{Al}=\frac{0,2.27}{11}.100\%=24,54\%\)
%m\(_{Fe}=100\%-24,54\%=75,46\%\)
b)ddX gồm H2SO4 dư , FeSO4, Al2(SO4)3
Theo pthh(1)(2)
n\(_{H2SO4}=n_{_{ }H2}=0,4\left(mol\right)\)
m dd(H2SO4)=\(\frac{0,4.98.100}{19,8}=197,98\%\)
C%(H2SO4)=\(\frac{0,4.98}{197,98}.100\%=19,8\%\)
Còn lại bạn tự tính nhé
