\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ PTHH:4K+O_2\underrightarrow{to}2K_2O\\ Vì:\dfrac{0,2}{4}< \dfrac{0,25}{1}\\ \rightarrow O_2dư.\\ n_{K_2O}=\dfrac{2}{4}.n_K=\dfrac{2}{4}.0,2=0,1\left(mol\right)\\ m_{K_2O}=94.0,1=9,4\left(g\right)\)