a/ \(2KMnO_4\left(x\right)\rightarrow K_2MnO_4\left(0,5x\right)+MnO_2\left(0,5x\right)+O_2\left(0,5x\right)\)
Gọi số mol của KMnO4 tham gia phản ứng là x.
\(\Rightarrow m_{KMnO_4}=158x\left(g\right)\)
\(\Rightarrow m_{K_2MnO_4}=0,5x.197=98,5x\left(g\right)\)
\(\Rightarrow m_{MnO_2}=0,5x.87=43,5x\left(g\right)\)
\(\Rightarrow22,12-158x+98,5x+43,5x=21,26\)
\(\Leftrightarrow x=0,05375\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,05375.0,5.22,4=0,602\left(l\right)\)
b/ \(m_{KMnO_4\left(pứ\right)}=0,05375.158=8,4925\left(g\right)\)
\(\Rightarrow\%KMnO_4=\dfrac{8,4925}{22,12}.100\%=38,39\%\)
c/ \(2HgO\left(0,05375\right)\rightarrow2Hg+O_2\left(0,026875\right)\)
\(\Rightarrow m_{pứ}=0,05375.201=10,80375\left(g\right)\)
Khối lượng HgO cần là: \(\dfrac{10,80375}{80\%}\approx13,505\left(g\right)\)