\(n_{O_2}=\dfrac{16.8}{22.4}=0.75\left(mol\right)\)
\(2KMnO_4\underrightarrow{^{t^0}}K_2MnO_4+MnO_2+O_2\)
\(1.5...............................................0.75\)
\(m_{KMnO_4}=1.5\cdot158=237\left(g\right)\)
\(n_{O_2} = \dfrac{16,8}{22,4} = 0,75(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 1,5(mol)\\ m_{KMnO_4} = 1,5.158 = 237(gam)\)
2KmnO4---> K2MnO4 + MnO2+ O2 (1)
ADCT n = V /22,4
nO2 = 16,8 /22,4=0,75(mol)
Theo pt(1) có
nO2/nKMnO4= 1/2 --> nKMnO4= 1/2 x 0,75= 0,375(mol)
ADCT
m=n.M
mKMnO4= 0,375 x 158= 59,25 (g)
