a, CuO+ 2HCl--> CuCl2 + H2O
Ta có nHCl=36,5.5/(100.36,5)=0,05 mol
=> nCuO=nHCl/2=0,025 mol
=> %mCuO=0,025.80.100/3,4=58,82%
%mCu=100-58,82=41,18%
b, HCl + NaOH--> NaCl + H2O
Ta có nHCl=0,05 mol
=> nH=0,05 mol
Trong PỨ trung hòa , nH=nOH=0,05 mol=nNaOH
=> mddNaOH=0,05.40.100/12=50/3g
=> VddNaOH=m/d=(50/3)/1,15=14,49 lít