nH2=\(\frac{1,12}{22,4}\)=0,05 mol
a. Fe+2HCl\(\rightarrow\)FeCl2+H2
nH2=nFe=0,05
\(\rightarrow\)mFe=0,05.56=2,8g
\(\rightarrow\)mFe2O3=6-2,8=3,2g
b. Fe2O3+6HCl\(\rightarrow\)2FeCl3+3H2O
nFe2O3=\(\frac{3,2}{160}\)=0,02
Ta có nHCl=2nFe+3nFe2O3=0,16
\(\rightarrow\)V HCl 2M=\(\frac{0,16}{2}\)=0,08l=80ml
c. nHCl=0,16; nNaOH=0,1
HCl+NaOH\(\rightarrow\)NaCl+H2O
nHCl>nNaOH\(\rightarrow\)HCl dư, NaOH hết
dd làm quỳ tím hóa đỏ