What is the mass of the precipitate formed during the interaction of an excess of sodium hydroxide solution

| September 24, 2021 | education

What is the mass of the precipitate formed during the interaction of an excess of sodium hydroxide solution with a solution containing 25 g of copper sulfate (1.1)

Let’s write the reaction equation:
2NaOH + CuSO4 = Cu (OH) 2 ↓ + Na2So4.
It can be seen from the reaction equation that the amount of substance is equal:
ν (CuSO4) = ν (Cu (OH) 2).
Knowing that ν (in-va) = m (in-va) / M (in-va), we have:
m (CuSO4) / M (CuSO4) = m (Cu (OH) 2) / M (Cu (OH) 2).
m (Cu (OH) 2) = m (CuSO4) / * M (Cu (OH) 2) / M (CuSO4).
Let’s define the molar masses:
M (CuSO4) = 63.5 + 32 + 16 * 4 = 159.5 g / mol.
M (Cu (OH) 2) = 63.5 + (16 + 1) * 2 = 97.5 g / mol.
Substituting the numerical values, we get:
m (Cu (OH) 2) = m (CuSO4) * M (Cu (OH) 2) / M (CuSO4) = 25 * 97.5 / 159.5 = 15.28 g.
Answer: the mass of the sediment is 15.28 g.



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