\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ CuO+H_2\underrightarrow{^{to}}Cu+H_2O\\ Vì:\dfrac{0,3}{1}>\dfrac{0,2}{1}\\ \Rightarrow H_2dư\\ n_{Cu\left(LT\right)}=n_{CuO}=0,2\left(mol\right)\\ n_{Cu\left(TT\right)}=\dfrac{8,96}{64}=0,14\left(mol\right)\\ H=\dfrac{0,14}{0,2}.100=70\%\)