[hóa 12] bài hóa hay

[thupham22011998]

Câu 16:
nHCl=0,2mol; nFeCl3=0,32 mol
X-NH2+HCl-->X-NH3Cl
3X-NH2+FeCl3+3H2O-->Fe(OH)3+3X-NH3Cl

Theo pthh, ta có: n X-NH2=0,2+0,32.3=1,16 mol -->m=1,16.17,25.2=40,02g