Đáp án+Giải thích các bước giải:

`M_{NaNO_3} = 23 + 14 + 16xx3 = 85` $(g/mol)$

`%Na = {23}/{85}xx100%` $\approx$ `27,1%`

`%N = {14}/{85}xx100%` $\approx$ `16,5%`

`%O = 100% - 27,1% - 16,5% = 56,4%`

`M_{K_2CO_3} = 39xx2 + 12 + 16xx3 = 138` $(g/mol)$`

`%K = {39.2}/{138}xx100%` $\approx$ `56,5%`

`%C = {12}/{138}xx100%` $\approx$ `8,7%`

`%O = 100% - 8,7% - 56,5% = 34,8%`

`M_{Al(OH)_3} = 27 + (16 + 1)xx3 = 78` $(g/mol)$

`%Al = {27}/{78}xx100%` $\approx$ `34,6%`

`%O = {16.3}/{78}xx100%` $\approx$ `61,5%`

`%H = 100% - 34,6% - 61,5% = 3,9%`

`M_{SO_2} = 32 + 16xx2 = 64` $(g/mol)$

`%S = {32}/{64}xx100% = 50%`

`%O = 100% - 50% = 50%`

`M_{SO_3} = 32 + 16xx3 = 80` $(g/mol)$

`%S = {32}/{80}xx100% = 40%`

`%O = 100% - 40% = 60%`

`M_{NH_4NO_3} = 14 + 1xx4 + 14 + 16xx3 = 80` $(g/mol)$

`%N = {14.2}/{80}xx100% = 35%`

`%H = {1.4}/{80}xx100% = 5%`

`%O = 100% - 35% - 5% = 60%`

`M_{Fe_2O_3} = 56xx2 + 16xx3 = 160` $(g/mol)$

`%Fe = {56.2}/{160}xx100% = 70%`

`%O = 100% - 70% = 30%` 

Giải thích các bước giải:

- Trong NaNO3 :

%Na = 23/85 .100% = 27,06%

%N = 14/85 .100% = 16,47%

%O = 16.3/85 .100% = 56,47%

- Trong K2CO3 :

%K = 39.2/138 .100% = 56,52%

%C = 12/138 . 100% = 8,7 %

%O = 16.3/138  .100% = 34,78%

- Trong Al(OH)3 :

%Al = 27/78  .100% = 34,62%

%O = 16.3/78  .100% = 61,54%

%H = 3/78  .100% = 3,84%

- Trong SO2 :

%S = 32/64 . 100% = 50%

%O = 100%-50% = 50%

- Trong SO3

%S = 32/80  .100% = 40%

%O = 100%-40% = 60%

- Trong Fe2O3 : 

%Fe = 56.2/160 .100% = 70%

%O = 100%-70% = 30%