PTHH: \(2CH_3COOH+CuO\rightarrow Cu\left(CH_3COO\right)_2+H_20\)
Ta có: \(n_{CuO}=\frac{40}{80}=0,5\left(mol\right)\)
Theo PTHH: \(n_{CH_3COOH}:n_{CuO}=2:1\)
\(\Rightarrow n_{CH_3COOH}=1mol\) \(\Rightarrow V_{CH_3COOH}=\frac{1}{2}=0,5\left(l\right)\)