Na2SO4+BaCl2=BaSO4+2NaCl0.16=m/0.1m=1.6г(BaCl2)n(BaCl2)=1.6/208=0.0077мольm(Na2SO4)=0.0077*126=0.97гc=m/V0.5=0.97/VV=1.94л=1940мл