Gọi a,b lần lượt là số mol của Al,Mg có trong hỗn hợp ban đầu
\(n_{khí}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
..a............3a............a.............1,5a......(mol)
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
..b...........2b..............b...........b............(mol)
Ta có hệ phương trình: \(\left\{{}\begin{matrix}27a+24b=6,3\\1,5a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,15\end{matrix}\right.\left(mol\right)\)
\(m_{AlCl_3}=133,5.0,1=13,35\left(g\right)\); \(m_{MgCl_2}=95.0,15=14,25\left(g\right)\)
nHCl = 3a+2b = 2(1,5a+b) = 2.0,3 = 0,6 (mol)
\(\Rightarrow V_{HCl}=\dfrac{0,6}{0,4}.1000=1500\left(ml\right)\)\(\Rightarrow m_{ddHCl}=V_{HCl}.d=1500.1,2=1800\left(g\right)\)
\(m_{ddsauPƯ}=m_{hh}+m_{ddHCl}-m_{H_2}\)= 6,3 + 1800 - 2.0,3 = 1805,7 (g)
\(C\%_{AlCl_3}=\dfrac{13,35}{1805,7}.100\approx0,74\%\)
\(C\%_{MgCl_2}=\dfrac{14,35}{1805,7}.100\approx0,79\%\)
