a/ Đặt nCu = x mol,  nCuO = y mol
n NO = 0.672/22.4 = 0.03 mol
ta có 64x + 80y = 3
BTe 2x = 3.0.03
=> x= 0.045 mol => mCu= 0.045. 64= 2.88 (g) =>%mCu = 2.88/3 . 100% =96%
     y= 0.0015 mol => m CuO = 0.12 (g) => %mCuO = 4%
b/ Bt mol Cu: nCu(NO2)2 =  0.045+ 0.0015 = 0.0465 mol
 => [Cu(NO3)2] = 0.0465 / 1.5 = 0.031 M
Có nHNO3 bđ = 0,1 . 1,5 =0,15 mol
BT mol N => nHNO3 dư = nHNO3 bđ - nNO - 2nCu(NO3)2 = 0,027 mol
[HNO3}= 0.027/1.5 =0,018 M