Câu trả lời:
a, PTHH
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
_x_____3x_______x______1,5x
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
_y_____2y_______y_______y
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
ta có hệ\(\left\{{}\begin{matrix}27x+24y=6,3\\1,5x+y=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,15\left(mol\right)\end{matrix}\right.\)
Vậy: \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1\times27}{6,3}=42,86\%\\\%m_{Mg}=100\%-42,86\%=57,14\%\end{matrix}\right.\)
b,\(n_{HCl}=3x+2y=0,6\left(mol\right)\)
thể tích dung dịch HCl cần dùng là:
\(V_{HCl}=\dfrac{n}{C_M}=\dfrac{0,6}{0,4}=1,5\left(l\right)=1500\left(ml\right)\)