\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\
n_{HCl}=1.0,25=0,25\left(mol\right)\\
pthh:Fe+2HCl\rightarrow FeCl_2+H_2\\
LTL:\dfrac{0,1}{1}< \dfrac{0,25}{2}\)
=> HCl dư
=> Fe tan hết
a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\); \(n_{HCl}=0,25.1=0,25\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,25}{2}\) => Fe hết, HCl dư
c)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,1---------------------->0,1
=> V = 0,1.22,4 = 2,24 (l)