\(n_{Mg}=a\left(mol\right),n_{Al}=b\left(mol\right)\)
\(m_{hh}=24a+27b=11.2\left(g\right)\)
\(n_{H_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(n_{H_2}=a+1.5b=0.4\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=\dfrac{2}{3},b=\dfrac{-8}{45}\)
Em xem lại đề nha !