\(n_{KMnO_4}=\dfrac{22,12}{158}=0,14\left(mol\right)\)
Gọi a là số mol KMnO4 pư
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
a---------------->0,5a----->0,5a
=> (0,14-a).158 + 197.0,5a + 87.0,5a = 21,16
=> a = 0,06 (mol)
=> \(\left\{{}\begin{matrix}n_{KMnO_4\left(dư\right)}=0,08\\n_{K_2MnO_4}=0,03\\n_{MnO_2}=0,03\end{matrix}\right.\)
PTHH: 2KMnO4 + 16HCl --> 2KCl + 2MnCl2 + 5Cl2 + 8H2O
0,08------>0,64
K2MnO4 + 8HCl --> 2KCl + MnCl2 + 2Cl2 + 4H2O
0,03------->0,24
MnO2 + 4HCl --> MnCl2 + Cl2 + 2H2O
0,03---->0,12
=> nHCl = 1 (mol)
=> mHCl = 1.36,5 = 36,5 (g)
=> \(m_{ddHCl}=\dfrac{36,5.100}{36,5}=100\left(g\right)\)
=> \(V_{ddHCl}=\dfrac{100}{1,18}=84,746\left(ml\right)\)
