\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{HCl}=0,1.1=0,1\left(mol\right)\)
Pt: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,1 -----> 0,05 -----> 0,05
Lập tỉ số : \(n_{Fe}:n_{HCl}=0,1>0,05\)
=> Fe dư, HCl hết
\(V_{H_2}=0,05.22,4=1,12\left(l\right)\)
\(n_{Fe\left(dư\right)}=0,1-0,05=0,05\left(mol\right)\)
\(m_{Fe\left(dư\right)}=0,05.56=2,8\left(g\right)\)