Đáp án:

a,

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

Bảo toàn khối lượng:

$m_{O_2}=m_{KMnO_4}-m_{\text{rắn}}=22,12-21,26=0,86$ gam 

$\Rightarrow n_{O_2}=\dfrac{0,86}{32}=0,026875$ mol 

$\Rightarrow V_{O_2}=0,026875.22,4=0,602L$ 

b,

$n_{KMnO_4\text{pứ}}=2n_{O_2}=0,026875.2=0,05375$ mol  

$\Rightarrow \%m_{KMnO_4\text{pứ}}=\dfrac{0,05375.158.100\%}{22,12}=38,39\%$ 

c,

$2HgO\xrightarrow{{t^o}} 2Hg+O_2$ 

$\Rightarrow n_{HgO}=2n_{O_2}=0,026875.2=0,05375$ mol 

$\Rightarrow m_{HgO\text{pứ}}=0,05375.217=11,66375$ gam 

Vậy $m_{HgO\text{cần dùng}}=11,66375:80\%=14,58$ gam