a/ \(2Cu+O_2\underrightarrow{t^o}2CuO\)
b/ \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
Theo PT => \(n_{O_2}=\dfrac{1}{2}n_{CuO}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
=> \(V_{O_2}=0,05.22,4=1,12\left(l\right)\)
c/ nCu =nCuO =0,1(mol)
=> mCu = 0,1 . 64 = 6,4 (gam)
