Na2SO4+BaCl2→BaSO4+2NaCl
mNa2SO4=170.25/100=42,5g
nNa2SO4=42,5/142=0,3mol
mBaCl2=208.20/100=41,6g
nBaCl2=41,6/208=0,2mol
Vì pứ theo tỉ lệ 1:1
Mà nNa2SO4>nBaCl2
→Na2SO4 dư, tính theo BaCl2
nBaSO4=nBaCl2=0,2mol
mBaSO4=0,2.233=46,6g
mdd sau pứ=170+208-46,6=331,4g
mNaCl=0,4.58,5=23,4g
C%NaCl=23,4/331,4.100%=7,06%
mNa2SO4 dư=0,1.142=14,2g
C%Na2SO4 dư=14,2/331,4.100%=4,28%