\(Fe+2HCl\xrightarrow[]{}FeCl_2+H_2\)
(mol) x..........2x..........x...........x
\(2Al+6HCl\xrightarrow[]{}2AlCl_3+3H_2\)
(mol) y...........3y............y............1,5y
a) Gọi x, y là số mol Fe, Al
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Ta có hpt: \(\left\{{}\begin{matrix}56x+27y=8,3\\x+1,5y=0,25\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\%^m_{Fe}=\dfrac{0,1.56}{8,3}.100=67,47\%\)
\(\%^m_{Al}=100-67,47=32,53\%\)
b) \(m_{FeCl_2}=0,1.127=12,7\left(g\right)\)
\(m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)
c) \(m_{ddHCl}=500.1,25=625\left(g\right)\)
d) Theo BTKL: \(m_{hh}+m_{ddHCl}=m_{ddA}+m_{H_2}\)
⇒ \(m_{ddA}=8,3+625-0,25.2=632,8\left(g\right)\)
\(C\%_{FeCl_2}=\dfrac{12,7}{632,8}.100=2\%\)
\(C\%_{AlCl_3}=\dfrac{13,35}{632,8}.100=2,1\%\)