Рассчитайте количество вещества сульфата бария, если в реакции с избытком сульфата натрия было использовано 4,16г. хлорида бария.
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20.05.2019 00:10
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2
BaCl2+Na2SO4=BaSO4 + 2NaCl
1)n(BaCl2)=4.16г/208г/моль=0.02моль
2)m(BaSO4)=0.02моль*233г/моль=4.66г
лучшее решение мне поставь и
BaCl2+Na2SO4-BaSO4осадок+2NaCl
1) n(BaCl2)=m/M=0,02 моль
2)По УХР: n(BaCl2)=n(BaSO4)=0,02моль