BaCl2+Na2SO4=BaSO4осадок+2NaCL
n( Na2SO4 )=36г/142г/моль=0.25моль
n ( BaSO4 ) = 0.25моль
m ( BaSO4 ) = 0.25моль*233г/моль=58.25г
m ( BaSO4 ) = 58.25г+50г=108.25г
Wвыход( BaSO4)=58.25г/108.25г*100%=53.8%
Copyright © 2025 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
BaCl2+Na2SO4=BaSO4осадок+2NaCL
n( Na2SO4 )=36г/142г/моль=0.25моль
n ( BaSO4 ) = 0.25моль
m ( BaSO4 ) = 0.25моль*233г/моль=58.25г
m ( BaSO4 ) = 58.25г+50г=108.25г
Wвыход( BaSO4)=58.25г/108.25г*100%=53.8%