\(\text{nFe = mFe : M Fe = 11,2 : 56 = 0,2 (mol)}\)
\(\text{PTHH: Fe + 2HCl }\rightarrow\text{FeCl2 + H2}\)
Theo PTHH: nFeCl2 = nFe = 0,2 (mol)
\(\rightarrow\) mFeCl2 = nFeCl2 .M FeCl2
= \(\text{0,2.127 = 25,4 }\)(g)
Theo PTHH: \(\text{nH2 = nFe = 0,2 (mol) }\)
\(\rightarrow\)\(\text{VH2(đktc) = 0,2.22,4 = 4,48 (l)}\)