\(2K+\frac{1}{2}O2\underrightarrow{^{to}}K2O\)(1)
\(K2O+2HCl\rightarrow2KCl+H2O\)(2)
Đổi 200ml = 0,2l
Ta có : \(nHCl=0,4.0,2=0,08\left(mol\right)\)
Theo PTHH(2):\(nK2O=\frac{1}{2}nHCl=\frac{1}{2}.0,08=0,04\left(mol\right)\)
Theo PT (1):\(nK=2nK2O=2.0,04=0,08\left(mol\right)\)
\(\rightarrow mK=0,08.39=3,12g\)
Theo PT (1) \(nO2=\frac{1}{2}nK2O=0,02\left(mol\right)\)
\(\rightarrow VO2=0,02.22,4=0,448ml\)
\(nKCl=\frac{1}{2}nHCl=0,08\)
\(\rightarrow mKCl=0,08.74,5=5,96g\)