nS = 8/32 = 0,25 (mol)

nFe2(SO4)3 = 0,25/3 = 1/12 (mol)

=> nO2 = 1/12 (mol)

PTHH: 2KMnO4 -> (t°) K2MnO4 + MnO2 + O2

nKMnO4 = 1/12 . 2 = 1/6 (mol)

mKMnO4 = 1/6 . 158 = 79/3 (g)

a)

Gọi $n_{KMnO_4} = a(mol) \Rightarrow n_{KClO_3} = 2a(mol)$

Suy ra : 

$158a + 122,5.2a = 40,3 \Rightarrow a = 0,1(mol)$

$m_{KMnO_4} = 0,1.158 = 15,8(gam)$
$m_{KClO_3} = 0,1.122,5 = 12,25(gam)$

b)

$2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2$

$2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$

Theo PTHH : 

$n_{O_2} = \dfrac{1}{2}n_{KMnO_4} + \dfrac{3}{2}n_{KClO_3} = 0,35(mol)$
$m_{O_2} = 0,35.32 = 11,2(gam)$

NTK Mn = \(158-39.1-4.16=55\left(đvC\right)\)

Đáp án : C

C₆H₅-CH = CH₂ + KMnO₄  ->  C₆H₅COOK + K₂CO₃ + MnO₂ + KOH + H₂O

a) Gọi số mol KMnO₄, KClO₃ là a, b (mol)

=> \(\left\{{}\begin{matrix}n_K=a+b\left(mol\right)\\n_{Mn}=a\left(mol\right)\\n_{Cl}=b\left(mol\right)\\n_O=4a+3b\left(mol\right)\end{matrix}\right.\)

Có \(n_O=\dfrac{9}{14}\Sigma_n\)

=> \(4a+3b=\dfrac{9}{14}\left(a+b+a+b+4a+3b\right)\)

=> \(\dfrac{1}{7}a-\dfrac{3}{14}b=0\) (1)

PTHH: 2KMnO₄ --t^o--> K₂MnO₄ + MnO₂ + O₂

                 a------------------------------->0,5a

             2KClO₃ --t^o--> 2KCl + 3O₂

                 b------------------>1,5b

=> \(0,5a+1,5b=\dfrac{10,08}{22,4}=0,45\) (2)

(1)(2) => a = 0,3 (mol); b = 0,2 (mol)

=> m = 0,3.158 + 0,2.122,5 = 71,9 (g)

b) \(\left\{{}\begin{matrix}\%m_{KMnO_4}=\dfrac{0,3.158}{71,9}.100\%=65,925\%\\\%m_{KClO_3}=\dfrac{0,2.122,5}{71,9}.100\%=34,075\%\end{matrix}\right.\)