2H2O \(\underrightarrow{to}\) 2H2 + O2 (1)
O2 + 4K \(\underrightarrow{to}\) 2K2O (2)
\(n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\)
Theo PT1: \(n_{O_2}=\dfrac{1}{2}n_{H_2O}=\dfrac{1}{2}\times0,2=0,1\left(mol\right)\)
Ta có: \(n_{O_2}\left(2\right)=n_{O_2}\left(1\right)=0,1\left(mol\right)\)
\(n_K=\dfrac{15,8}{39}=\dfrac{79}{195}\left(mol\right)\)
Theo PT2: \(n_{O_2}=\dfrac{1}{4}n_K\)
Theo bài: \(n_{O_2}=\dfrac{39}{158}n_K\)
Vì \(\dfrac{39}{158}< \dfrac{1}{4}\) ⇒ K dư, O2 hết ⇒ Tính theo O2
Theo PT2: \(n_{K_2O}=2n_{O_2}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{K_2O}=0,2\times94=18,8\left(g\right)\)
Vậy \(m=18,8\left(g\right)\)
