\(PTPƯ:CuO+H_2\rightarrow Cu=H_2O\)
\(nCuO=\dfrac{48}{80}=0,6mol\)
\(Theo\) \(pt:\) \(nH_2=nCuO=0,6mol\)
\(\Rightarrow VH_2=0,6.22,4=13,44lít\)
\(Theo\) \(pt:\) \(nCu=nCuO=0,6mol\)
\(\Rightarrow mCu=0,6.64=38,4g\)
\(\Rightarrow12.B\\ \Rightarrow13.A\)