3.Na2SO4+BaCl2→BaSO4+2NaCl
nNa2SO4=2*01=0,2mol
nBaCl2=0,5*0,2=0,1mol
Vậy Na2SO4 dư 0,1mol
nBaSO4=nBaCl2=0,1mol
m tủa=0,1*233=23,3g
4.CuO+2HCl→CuCl2+H2O
nCuO=8/80=0,1mol
mHCl=100*36,5/100=36,5g
nHCl=36,5/36,5=1mol
nHCl phản ứng=2nCuO=0,2mol
⇒HCl dư 0,8mol
mCuCl2=0,1*135=13,5g
mdd sau phản ứng=8+100=108g
C%CuCl2=13,5/108*100%=12,5%
C%HCl dư=0,8*36,5/108*100=27,04%