nFe2O3=16/160=0.1 (mol)
mHCl=146*20/100=29.2(g)
=> nHCl=29.2/36.5=0.8(mol)
a,PTHH:Fe2O3 + 6HCl => 2FeCl3 + 3H2O
Ban đầu: 0.1 0.8
PƯ: 0.1 0.6 0.2 0.3 (mol)
Sau PƯ: 0 0.2 0.2 0.3 (mol)
b, theo PTHH: nHCl sau pư = 0.2 (mol) => nHCl= 0.2*36.5=7.3 (g)
=> C%= 7.3/146 *100=5%