H2+CuO-to>Cu+H2O
0,1----0,1----0,1
n H2=0,1 mol
n CuO=0,15 mol
=>m Cu=0,1.64=6,4g
=>m CuO=0,05.80=4g
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\\ pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ LTL:\dfrac{0,1}{1}< \dfrac{0,15}{1}\\ =>CuO\left(d\right)\\ n_{CuO\left(pu\right)}=n_{Cu}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_{Cu}=0,1.64=6,4\left(g\right)\\m_{CuO\left(d\right)}=\left(0,15-0,1\right).80=4\end{matrix}\right.\\ =>B\)