giúp e với ạ

cảm ơn

`a)` `@` `Fe(NO_3)_3:`

             `\text{Ta có:}` `%m_(Fe) = 56/242 . 100% ≈ 23,14%`

             `\text{Ta có:}` `%m_(N) = 42/242 . 100% ≈ 17,36%`

             `\text{Ta có:}` `%m_(O) = 144/242 . 100% ≈ 59,5%`

    `@` `Fe(NO_3)_2:`

             `\text{Ta có:}` `%m_(Fe) = 56/180 . 100% ≈ 31,11%`

             `\text{Ta có:}` `%m_(N) = 28/180 . 100% ≈ 15,56%`

             `\text{Ta có:}` `%m_(O) = 96/180 . 100% ≈ 53,33%`

`b)` `@` `N_2O:`

             `\text{Ta có:}` `%m_(N) = 28/44 . 100% ≈ 63,64%`

             `\text{Ta có:}` `%m_(O) = 16/44 . 100% ≈ 36,36%`

    `@` `NO:`

             `\text{Ta có:}` `%m_(N) = 14/30 . 100% ≈ 46,67%`

             `\text{Ta có:}` `%m_(O) = 16/30 . 100% ≈ 53,33%`

    `@` `NO_2:`

             `\text{Ta có:}` `%m_(N) = 14/46 . 100% ≈ 30,43%`

             `\text{Ta có:}` `%m_(O) = 32/46 . 100% ≈ 69,57%`

Đáp án:

Giải thích các bước giải:

 a) Fe($NO_3$)

%$M_{Fe}$= $\frac{56}{56+14+16×3}$ ×100% = 47,46%

%$M_{N}$ =$\frac{14}{56+14+16×3}$ ×100% = 11,86%

%$M_{O}$ = 100% - ( 47,46% + 11,86%)=40,68%

Fe($NO_3$)$_{2}$ 

%$M_{O}$=$\frac{16×3}{56+14×3+16×9}$ ×100%  = 11%

%$M_{Fe}$=$\frac{56}{56+14×3+16×9}$ ×100% = 38,37%

%$M_{N}$=100%-(11%+38,37%) = 50,63%

b) $N_{2}$O

%$M_{O}$=$\frac{16}{14×2+16}$ ×100% = 36% %

$M_{N}$= 100% - 36% = 64%

NO

%$M_{O}$=$\frac{16}{14+16}$ ×100%  = 54,4%

%$M_{N}$=100%- 54,4%=45,6%

N$O_{2}$ %$M_{N}$=$\frac{14}{16×2+14}$ ×100%  = 30%

%$M_{O}$=100% - 30% = 7-%