a) Fe+2HCl---->FeCl2+H2
b) n\(_{Fe}=\frac{22,4}{56}=0,4\left(mol\right)\)
n\(_{HCl}=\frac{18,25}{36,5}=0,5\left(mol\right)\)
Lập tỉ lệ ta thấy HCl dư
Theo pthh
n\(_{HCl}=2n_{Fe}=0,4\left(mol\right)\)
n\(_{Fe}dư=0,5-0,4=0,1\left(mol\right)\)
m\(_{HCl}dư=0,1.36,5=3,65\left(g\right)\)
c) Theo pthh
n\(_{H2}=n_{Fe}=0,2\left(mol\right)\)
V\(_{H2}=0,2.2,24=4,48\left(l\right)\)