\(a) 2Cu + O_2 \xrightarrow{t^o} 2CuO\\ n_{Cu} = \dfrac{51,2}{64} = 0,8(mol)\\ n_{O_2} = \dfrac{1}{2}n_{Cu} = 0,4(mol) \Rightarrow V_{O_2} = 0,4.22,4 = 8,96(lít)\\ b) n_{Cu\ pư} = n_{CuO} = \dfrac{48}{80} = 0,6(mol)\\ \Rightarrow H = \dfrac{0,6}{0,8}.100\% = 75\%\)
