$pthh :$
$2KMnO4→K2MnO4+MnO2+O2$
Áp dụng $ĐLBTKL :$
$⇒m_{giảm}=m_{O2}=79-72,6=6,4g$
$⇒n_{O2}=6,4/32=0,2mol$
theo pt :
$n_{KMnO4 pư }=2.n_{O2}=0,4mol$
$⇒m_{KMnO4 pư }=0,4.158=63,2g$
$n_{K2MnO4}=n_{O2}=0,2mol$
$⇒m_{K2MnO4}=0,2.197=39,4g$
$n_{MnO2}=n_{O2}=0,2mol$
$⇒m_{MnO2}=0,2.87=17,4g$
⇒KMnO4 dư :
$m_{KMnO_{4}dư}=79-63,2=15,8g$
$⇒\%m_{KMnO4}=\frac{15,8.100\%}{72,6}=21,76\%$
$\%m_{K2MnO4}=\frac{39,4.100\%}{72,6}=54,27\%$
$\%{MnO2}=100\%-21,76\%-54,27\%=23,97\%$
$b/$
$H\%=\frac{63,2.100}{79}=80\%$