a) PTHH: Fe(r) + 2HCl(dd) -> FeCl2(dd) + H2(k)
b) nH2= 1,12:22,4=0,05(mol)
theo pt ở ý a ta có: nH2=nFe=0,05(mol)
-> mFe=0,05.56=2,8(g)
c) Theo pt ý a ta có: nHCl=2nFe=0,05.2=0,1(mol)
-> mHCl=0,1.36,5=3,65(g)
ADCT: C%=mct/mdd.100%
-> mdd=mct/C%.100%
-> mddHCl=mHCl/10%.100%=36,5(g)