Đáp án:

 dưới nha bn

Giải thích các bước giải:

     2KMnO4→K2MnO4+MnO2+O2

    2,125         1,0625  1,0625  ←1,0625

m O2=500-466=34 g 

a.H=$\frac{2,125 .158}{500}$ x100=67,15%

b. V O2(đktc)=1,0625 x22,4=23,8 l

c.%m MnO2=$\frac{1,0625 .87}{466}$ x100≈19,8%

%m K2MnO4=80,2%

.....................chúc bn hk tốt.......................

$n_{KMnO_4}=500/158=3,165mol$

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

a.Theo pt :

$n_{K_2MnO_4}=n_{MnO_2}=1/2.n_{KMnO_4}=1/2.3,165=1,5825mol$

$⇒m_{K_2MnO_4}=1,5825.197=311,7525g$

$m_{MnO_2}=1,5825.87=137,6675g$

$⇒m_{rắn}=311,7525+137,6675=449,43$

$⇒H\%=\dfrac{449,43}{466}.100\%=96,44\%$

b.Theo pt :

$n_{O_2}=1/2.n_{KMnO_4}=1/2.3,165=1,5825mol$

$⇒V_{O_2}=1,5825.22,4=35,448l$

$c.\%m_{KMnO_4}=\dfrac{311,7525}{466}.100\%≈67\%$

$\%m_{MnO_2}=100\%-67\%=33\%$