a,
\(n_{H2}=0,6\left(mol\right)\)
Gọi a là mol Al, b là mol Fe
\(\Rightarrow27a+56b=19,5\left(1\right)\)
Bảo toàn e: \(3a+2b=0,6.2=1,2\left(2\right)\)
(1)(2) \(\Rightarrow\left\{{}\begin{matrix}a=0,25\\b=0,23\end{matrix}\right.\)
\(\%_{Al}=\frac{0,25.27.100}{19,5}=34,6\%\)
\(\%_{Fe}=100\%-34,6\%=65,4\%\)ư
b,
\(n_{Cl}=n_{HCl}=2n_{H2}=1,2\left(mol\right)\)
\(\Rightarrow m_{Cl}=42,6\left(g\right)\)
\(m_{muoi_{clorua}}=19,5+42,6=62,1\left(g\right)\)