\(2M+O_2\rightarrow\left(t^o\right)2MO\\ m_{O_2}=m_{MO}-m_M=9,6-7,68=1,92\left(g\right)\\ \Rightarrow n_{O_2}=\dfrac{1,92}{32}=0,06\left(mol\right)\\ \Rightarrow n_M=0,06.2=0,12\left(mol\right)\\ \Rightarrow M_M=\dfrac{7,68}{0,12}=64\left(\dfrac{g}{mol}\right)\\ \Rightarrow M:Đồng\left(Cu=64\right);Oxit:Đồng\left(II\right)oxit\left(CuO\right)\)
=> C
Câu 5:
\(4M+O_2\rightarrow\left(t^o\right)2M_2O\\ m_{O_2}=m_{oxit}-m_M=2,82-0,34=0,48\left(g\right)\\ \Rightarrow n_{O_2}=\dfrac{0,48}{32}=0,015\left(mol\right)\\ \Rightarrow n_M=4.0,015=0,06\left(mol\right)\\ \Rightarrow M_M=\dfrac{2,34}{0,06}=39\left(\dfrac{g}{mol}\right)\\ \Rightarrow M:Kali\left(K=39\right);Oxit:K_2O\\ \Rightarrow B\)
