Bài 2:
Ta có: \(\left\{{}\begin{matrix}n_{Mg}=\dfrac{0,6}{24}=0,025\left(mol\right)\\n_{H_2\left(axit\:\right)}=\dfrac{1,568}{22,4}=0,07\left(mol\right)\end{matrix}\right.\)
PTHH: \(Al+NaOH+H_2O\rightarrow NaAlO_2+\dfrac{3}{2}H_2\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
0,025____________________0,025 (mol)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
0,03_______________________0,045 (mol)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,6}{0,6+0,03\cdot27}\cdot100\%\approx42,55\%\\\%m_{Al}=57,45\%\end{matrix}\right.\)