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`
