Vì Cu không phản ứng với HCl
\(\Rightarrow m_{Cu}=6,4\left(g\right)\Rightarrow m_{Al+Fe}=11,9-6,4=5,5\left(g\right)\)
\(n_{H_2}=\frac{4,48}{22,4}=0,2\left(mol\right)\Rightarrow m_{H_2}=0,2.2=0,4\left(g\right)\)
BTNT H: \(\Rightarrow n_{HCl}=2.n_{H_2}=2.0,2=0,4\left(mol\right)\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(BTKL:m_{Muoi}=m_{Fe+Al}+m_{HCl}-m_{H_2}=5,5+14,6-0,4=19,7\left(g\right)\)