Câu hỏi của Thuỳ Lê Minh

a) \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)PTHH: \(4P+5O_2\xrightarrow[]{t^o}2P_2O_5\) 0,4-->0,5----->0,2b) \(V_{O_2}=0,5.24,79=12,395\left(l\right)\)c) \(m_{P_2O_5}=0,2.142=28,4\left(g\right)\)d) PTHH: \(2Cu+O_2\xrightarrow[]{t^o}2CuO\) 1 m_{Cu} = 1.64 = 64 (g)`Câu trả lời: a) \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)PTHH: \(4P+5O_2\xrightarrow[]{t^o}2P_2O_5\) 0,4-->0,5----->0,2b) \(V_{O_2}=0,5.24,79=12,395\left(l\right)\)c) \(m_{P_2O_5}=0,2.142=28,4\left(g\right)\)d) PTHH: \(2Cu+O_2\xrightarrow[]{t^o}2CuO\) 1 m_{Cu} = 1.64 = 64 (g)`