a) \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\end{matrix}\right.\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Xét tỉ lệ: \(0,2>\dfrac{0,3}{2}\Rightarrow\) Fe dư
Theo PTHH: \(n_{Fe\left(p\text{ư}\right)}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)
\(\Rightarrow m_{Fe\left(d\text{ư}\right)}=\left(0,2-0,15\right).56=2,8\left(g\right)\)
c) \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)