a)2Al +3H2SO4---->Al2(SO4)3 +3H2
x(27x)---------------------------------1,5x
Fe+ H2SO4---->FeSO4+H2
y(56y)-------------------------y
b) n\(_{H2}=\frac{0,56}{22,4}=0,025\left(mol\right)\)
Gọi n Al=x
n Fe =y
Theo bài ta có hệ pt
\(\left\{{}\begin{matrix}27x+56y=0,83\\1,5x+y=0,025\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,01\\y=0,01\end{matrix}\right.\)
m\(_{Al}=0,01.27=0,27\left(g\right)\)
m\(_{Fe}=0,01.56=0,56\left(g\right)\)