Đáp án:

Giải thích các bước giải:

`ZnSO_4`

`M_{ZnSO_4}=65+32+16.4=161(g`$/$`mol)`

`%m_{Zn}=\frac{65}{161}.100=40,373%`

`%m_{S}=\frac{32}{161}.100=19,876%`

`%m_{O}=\frac{16.4}{161}.100=39,751%`

`-----------`

`ZnCl_2`

`M_{ZnCl_2}=65+35,5.2=136(g`$/$`mol)`

`%m_{Zn}=\frac{65}{136}.100=47,794%`

`%m_{Cl}=\frac{35,5.2}{136}.100=52,206%`

`-----------`

`Zn(OH)_2`

`M_{Zn(OH)_2}=65+2(16+1)=99(g`$/$`mol)`

`%m_{Zn}=\frac{65}{99}.100=65,657%`

`%m_{O}=\frac{16.2}{99}.100=32,323%`

`%m_{H}=\frac{1.2}{99}.100=2,02%`

`-----------`

`Fe_2(SO_4)_3`

`M_{Fe_2(SO_4)_3}=56.2+3(32+16.4)=400(g`$/$`mol)`

`%m_{Fe}=\frac{56.2}{400}.100=28%`

`%m_{S}=\frac{32.3}{400}.100=24%`

`%m_{O}=\frac{16.4.3}{400}.100=48%`

`------------`

`FeCl_3`

`M_{FeCl_3}=56+35,5.3=162,5(g`$/$`mol)`

`%m_{Fe}=\frac{56}{162,5}.100=34,461%`

`%m_{Cl}=\frac{35,5.3}{162,5}.100=65,539%`

`------------`

`Fe(OH)_3`

`M_{Fe(OH)_3}=56+3(16+1)=107(g`$/$`mol)`

`%m_{Fe}=\frac{56}{107}.100=52,336%`

`%m_{O}=\frac{16.3}{107}.100=44,86%`

`%m_{H}=\frac{1.3}{107}.100=2,804%`

`#Devil`