\(pthh:2Cu\left(NO_3\right)_2\overset{t^o}{--->}2CuO+4NO_2\uparrow+O_2\uparrow\)
Ta có: \(n_{Cu\left(NO_3\right)_2}=\dfrac{18,8}{188}=0,1\left(mol\right)\)
Theo pt: \(n_{CuO}=n_{Cu\left(NO_3\right)_2}=0,1\left(mol\right)\)
\(\Rightarrow m=m_{CuO}=0,1.80=8\left(g\right)\)