2KMnO4-to>K2MnO4+MnO2+O2
1,2-------------------------------------0,6 mol
n O2=13,44\22,4=0,6 mol
H =75%
=>m KMnO4 tt= 1,2.158 .100\75=252,8g
\(n_{O_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: 2KMnO4 ---to→ K2MnO4 + MnO2 + O2
Mol: 1,2 0,6
\(m_{KMnO_4\left(lt\right)}=1,2.158=189,6\left(g\right)\Rightarrow m_{KMnO_4\left(tt\right)}=\dfrac{189,6}{75}.100=252,8\left(g\right)\)
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
\(\Rightarrow n_{KMnO_4}=\dfrac{2n_{O_2}}{75\%}=\dfrac{2.13,44.100}{22,4.75}=1,6\left(mol\right)\)
\(\Rightarrow a=m_{KMnO_4}=1,6.158=252,8\left(g\right)\)
