Натрій карбонат масою 1.06г прореагував з кальцій хлоридом, що знаходився у розчині . Визначте масу осаду , що утворився

Объяснение:

Дано:

m(Na2CO3) = 1.06 г.

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m(CaCO3) - ?

Na2CO3 + CaCl2 = 2NaCl + CaCO3↓

M(Na2CO3) = 106 г/моль

n(Na2CO3) = 1.06/106 = 0.01 моль

n(CaCO3):n(Na2CO3) = 1:1

n(CaCO3) = 0.01 моль

M(CaCO3) = 100 г/моль

m(CaCO3) = 100 * 0.01 = 1 г

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Объяснение: