nCuO=3,2/80=0,04(mol
mH2SO4=200.20%=40(g) => nH2SO4=40/98=20/49(mol)
a) PTHH: CuO + H2SO4 -> CuSO4 + H2O
Ta có: 0,04/1 < 20/49:1
=> H2SO4 dư, CuO hết, tính theo nCuO
b) mddsau=mCuO+mddH2SO4=3,2+200=203,2(g)
nH2SO4(p.ứ)=nCuSO4=nCuO=0,04(mol)
=> mH2SO4(dư)= 40- 0,04.98=36,08(g)
mCuSO4=0,04.160=6,4(g)
=> C%ddCuSO4=(6,4/203,2).100=3,15%
C%ddH2SO4(dư)= (36,08/203,2).100=17,756%
CuO +2HCl= CuCl2 +H2O
ZnO+2HCl= ZnCl2 +H2O
gọi x,y là mol của CuO, ZnO
80x + 81y = 12.1
2x+2y = 0.3
=> x=0.05 , y=0.1 => mCuO= 4 %CuO=4/12.1 m ZnO=8.1 =>%ZnO=8.1/12.1
nH2SO4=1/2nHCl=0.3/2 =0.15
mH2SO4=0.15x98=14.7g => mddH2SO4=14.7/20%=73.5g
