$448cm^3=0,448(l);100(ml)=0,1(l)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ FeO+2HCl\to FeCl_2+H_2O\\ b,n_{H_2}=\dfrac{0,448}{22,4}=0,02(mol)\\ n_{HCl}=1,5.0,1=0,15(mol)\\ \text{Lập tỉ lệ: }\dfrac{n_{HCl}}{2}>\dfrac{n_{H_2}}{1}\Rightarrow HCl\text{ dư}\\ \Rightarrow n_{Fe}=n_{H_2}=0,02(mol)\\ \Rightarrow m_{Fe}=0,02.56=1,12(g)\\ \Rightarrow m_{FeO}=4-1,12=2,88(g)\\ c,n_{FeO}=\dfrac{2,88}{160}=0,018(mol)\\ \Rightarrow \Sigma n_{HCl(p/ứ)}=2n_{Fe}+2n_{FeO}=0,036+0,04=0,076(mol)\\ \Rightarrow n_{HCl(dư)}=0,15-0,076=0,074(mol)\\ \Sigma n_{FeCl_2}=n_{Fe}+n_{FeO}=0,038(mol)\\ \Rightarrow C_{M_{HCl(dư)}}=\dfrac{0,074}{0,1}=0,74M\\ C_{M_{FeCl_2}}=\dfrac{0,038}{0,1}=0,38M$