\(\text{nH2 = 0,5 mol}\)
PTHH: Fe + 2HCl \(\rightarrow\) FeCl2 +H2
______0,2___0,4_____0,2____0,2___(mol)
\(\text{x = mFe = 0,2. 56 = 11,2 gam}\)
mdd sau phản ứng = mFe + mdd HCl - mH2
\(\text{= 11,2 + 400 - 0,2. 2 }\)
= 410,8 gam
C% FeCl2 = \(\text{(0,2. 127. 100)(410,8)= 6,18%}\)