a) nCuO=1,680=0,02(mol)nCuO=1,680=0,02(mol)
nH2SO4=20%.10098=1049(mol)nH2SO4=20%.10098=1049(mol)
PTHH: CuO + H2SO4 --> CuSO4 + H2O
Xét tỉ lệ: 0,021<104910,021<10491 => CuO hết, H2SO4 dư
b)
PTHH: CuO + H2SO4 --> CuSO4 + H2O
0,02-->0,02------->0,02
=> mCuSO4=0,02.160=3,2(g)mCuSO4=0,02.160=3,2(g)
c) mdd sau pư = 1,6 + 100 = 101,6 (g)
C%H2SO4.dư=98.(1049−0,02)101,6.100%=17,756%