nFe= 5.6/56=0.1 mol
mHCl= 109.5*10/100=10.95g
nHCl= 10.95/36.5=0.3 mol
____Fe + 2HCl --> FeCl2 + H2
Bđ: 0.1___0.3
Pư: 0.1___0.2_____0.1____0.1
Kt: 0____0.1______0.1____0.1
VH2= 0.1*22.4=2.24l
mH2= 0.1*2=0.2g
mHCl dư= 0.1*36.5=3.36g
mFeCl2= 0.1*127=12.7g
mdd sau phản ứng=mFe + mddHCl - mH2= 5.6+109.5-0.2=114.9g
C%HCl dư= 3.65/114.9*100%= 3.17%
C%FeCl2= 12.7/114.9*100%= 11.05%