`a)PTHH: Fe_2O_3+6HCl->2FeCl_3+3H_2O`
`b)n_(Fe_2O_3)=m/M=(3,2)/(56.2+16.3)=0,02(mol)`
Theo `PTHH: n_(HCl)=6n_(Fe_2O_3)=6.0,02=0,12(mol)`
`=>m_(HCl)=n.M=0,12(1+35,5)=4,38(g)`
`=>C%_(HCl)=(mct)/(mdd).100%=(4,38)/(200).100%=2,19%`
`c)`Theo `PTHH: n_(FeCl_3)=2n_(Fe_2O_3)=2.0,02=0,04(mol)`
`=>m_(FeCl_3)=n.M=0,04(56+35,5.3)=6,5(g)`