a)\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
x______________________x
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
y________________________1,5y
b)\(n_{H_2}=\frac{1,568}{22,4}=0,07\left(mol\right)\)
Gọi x là nMg;y là nAl
Ta có hpt:
\(\begin{cases}24x+27y=1,41\\x+1,5y=0,07\end{cases}\Leftrightarrow\begin{cases}x=0,025\\y=0,03\end{cases}\)
Suy ra: mMg=0,025.24=0,6 (g)
=>\(\%m_{Mg}=\frac{0,6.100}{1,41}=42,55\%;\%m_{Al}=57,45\%\)