Theo gt ta có: \(n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\)
\(1Bar=0,9869atm\)
PTHH: \(2Cu+O_2\rightarrow2CuO\)
Ta có: \(n_{CuO}=n_{Cu}=0,2\left(mol\right)\Rightarrow a=m_{CuO}=16\left(g\right)\)
\(n_{O_2}=\dfrac{1}{2}.n_{Cu}=0,1\left(mol\right)\Rightarrow V=\dfrac{n.R.T}{p}=\dfrac{0,1.\dfrac{22,4}{273}.\left(273+20\right)}{0,9869}=2,436\left(l\right)\)