Fe + 2HCl -> FeCl2 + H2
1 2 1 1
nH2 = 4,48/22,4 = 0,2 (mol) => nFe = 0,2 mol
mFe = 0,2.56 = 11,2 (gam)
C%HCl = mHCl/mdd.100% = [(0,4.36,5)/250].100 = 5,84%
mdd = 11,2 + 250 - 0,2.1 = 261 (gam)
C%FeCl2 = mFeCl2/mdd.100% = [(0,2.127)/261].100 = 9,73%
Goodluck!!!