\(n_{H_2}=\dfrac{7.84}{22.4}=0.35\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
x 3x x 1,5x
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
y 2y y y
Theo đề, ta có hệ: 27x+56y=13,9 và 1,5x+y=0,35
=>x=0,1; y=0,2
\(m_{Al}=0.1\cdot27=2.7\left(g\right)\)
\(m_{Fe}=0.2\cdot56=11.2\left(g\right)\)
