2KMnO4-to>K2MnO4+MnO2+O2
0,2---------------------------------------0,1 mol
n O2=\(\dfrac{2,24}{22,4}\)=0,1 mol
=>m KMnO4 tt =0,2.158=31,6g
=>H =\(\dfrac{31,6}{39,5}.100\)=80%
Ta có: \(n_{KMnO_4}=\dfrac{39,5}{158}=0,25\left(mol\right)\)
PT: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)
Theo PT: \(n_{O_2\left(LT\right)}=\dfrac{1}{2}n_{KMnO_4}=0,125\left(mol\right)\)
Mà: \(n_{O_2\left(TT\right)}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(\Rightarrow H\%=\dfrac{0,1}{0,125}.100\%=80\%\)
Bạn tham khảo nhé!
\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Pthh 2KMnO4 → (to) K2MnO4 + MnO2 + O2↑
0,2 0,1
\(m_{KMnO_4\left(pu\right)}=0,2.158=31,6\left(g\right)\)
\(H\%=\dfrac{31,6}{39,5}=80\%\)
