Предмет: Химия, автор: SingerPurty

Сколько грамм осадка получится при взаимодействии серной кислоты с 52 грамма 20%-го раствора хлорида баррия

Ответы

Автор ответа: tomiriska200907

1

Ответ:

Дано:

m(BaCl2)=52

w(BaCl2)=20%

m(BaSO4)=?

Решение:

m(BaCl2)=52*0.2=10.4

10.4 x

BaCl2 + H2SO4 → BaSO4↓ + 2HCl

208 233

x=10.4*233/208=11.65грамм

Объяснение:

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