a) 2Al + 3H2SO4 → Al2(SO4)3 + 3H2 (1)
Mg + H2SO4 → MgSO4 + H2 (2)
b) \(n_{H_2}=\frac{3,92}{22,4}=0,175\left(mol\right)\)
Gọi x,y lần lượt là số mol của Al và Mg
Ta có: \(27x+24y=3,75\) (*)
Theo PT1: \(n_{H_2}=1,5n_{Al}=1,5x\left(mol\right)\)
Theo pT2: \(n_{H_2}=n_{Mg}=y\left(mol\right)\)
Ta có: \(1,5x+y=0,175\) (**)
Tù (*)(**) ta có: \(\left\{{}\begin{matrix}27x+24y=3,75\\1,5x+y=0,175\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,1\end{matrix}\right.\)
Vậy \(n_{Al}=0,05\left(mol\right)\Rightarrow m_{Al}=0,05\times27=1,35\left(g\right)\)
\(n_{Mg}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1\times24=2,4\left(g\right)\)
\(\%m_{Al}=\frac{1,35}{3,75}\times100\%=36\%\)
\(\%m_{Mg}=\frac{2,4}{3,75}\times100\%=64\%\)