Giúp mình với ạ mình cảm ơn

Đáp án: `+` Giải thích các bước giải:

 `a,` CTHH: \(\text{Al}_2\text{O}_3\)

`\%\text{Al} = \frac{2 . 27}{2 . 27 + 3 . 16} . 100\% \approx 52,94\%`

`\%\text{O} = 100\% - 52,94\% \approx 47,06\%`

`b,` CTHH: \(\text{KCl}\)

`\%\text{K} = \frac{39}{39 + 35,5} . 100\% \approx 52,35\%`

`\%\text{Cl} = 100% - 52,35% \approx 47,65\%`

`c,` CTHH: \(\text{Cu(OH)}_2\)

`\%\text{Cu} = \frac{64}{64 + 2 . (16 + 1)} . 100\% \approx 65,31\%`

`\%\text{O} = \frac{2 . 16}{98} . 100\% \approx 32,65\%`

`\%\text{H} = \frac{2 . 1}{98} . 100\% \approx 2,04\%`

`d,` CTHH: \(\text{Fe}_2(\text{SO}_4)_3\)

`\%\text{Fe} = \frac{2 . 56}{400} . 100\% = 28\%`

`\%\text{S} = \frac{3 . 32}{400} . 100\% = 24\%`

`\%\text{O} = 100% - 28% - 24% = 48\%`

`e,` CTHH: \(\text{Zn(NO}_3\text{)}_2\)

`\%\text{Zn} = \frac{65}{189} . 100\% \approx 34,39\%`

`\%\text{N} = \frac{2 . 14}{189} . 100\% \approx 14,81\%`

`\%\text{O} = 100% - 34,39% - 14,81% \approx50,80\%`