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При взаимодействии нитрата серебра с фосфатом калия было получено 0,2 моль фосфата серебра. Какая масса нитрата серебра вступила в реакцию?
Ответы 1
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3AgNO3 + K3PO4 → Ag3PO4↓ + 3KNO31) 3(AgNO3) — 1(Ag3PO4)n(AgNO3) ——— 0,2моль(Ag3PO4)По пропорции: n(AgNO3) = 0,2моль * 3 : 1 = 0,6моль2) m(AgNO3) = 0,6моль * 170г/моль = 102г
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Автор:
faustinosdxg - 5 лет назад
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Английский язык -
Автор:
marelistrong - 5 лет назад
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