\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ LTL:0,2>0,15\Rightarrow CuO.dư\\ Theo.pt:n_{CuO\left(pư\right)}=n_{Cu}=n_{H_2}=0,15\left(mol\right)\\ \Rightarrow m_{chất.rắn}=\left(0,2-0,15\right).80+64.0,15=13,6\left(g\right)\)