n(Fe)=0.56/56=0.01(mol)
n(HCl)=1*50*10^-3=0.05(mol)
(1)Fe+2HCl===FeCl2+H2↑
n(H2)=n(Fe)=0.01(mol)
V(H2)=Vm*n(H2)=22.4*10^3*0.01=224(mL)
(2)n(H(+))=n(HCl)-2n(Fe)=0.05-2*0.01=0.03(mol)
c(Fe(2+))=0.01/(50*10^-3)=0.2(mol/L)
c(H(+))=0.03/(50*10^-3)=0.6(mol/L)