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Автор - shpilevskaya20

решите пожалуйста 819 задачу.

Ответ

Автор - kamilakulieva
Na2SO4 + BaCl2 = BaSO4 + 2NaCl
m(Na2SO4)=7,6/100х200=15,2 г
M(Na2SO4)=142 г/моль
n(Na2SO4)=15,2/142=0,107 моль
n(Na2SO4)=n(BaSO4)=0,107 моль
M(BaCl2)=208 г/моль
m(BaCl2)=208х0,107=22,26 г

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