Question

Сколько граммов осадка образуется при взаимодействие 24г CuSO₄ с необходимом количеством NaOH? Очень нужно 25б

Answer 1

2NaOH+CuSO4->Cu(OH)2+Na2SO4
находим количество вещества CuSO4
n=24г./160г./моль=0.15 моль
n(Cu(OH)2)=n(CuSO4)
находим массу Cu(OH)2
m=0.15моль×98г./моль=14.7г.

Answer 2

Calculation of Precipitate Formation

To calculate the amount of precipitate formed when 24g of CuSO₄ reacts with the necessary amount of NaOH, we can use the following chemical equation:

CuSO₄ + 2NaOH → Cu(OH)₂ + Na₂SO₄

According to the equation, 1 mole of CuSO₄ reacts with 2 moles of NaOH to produce 1 mole of Cu(OH)₂. The molar mass of CuSO₄ is 159.609 g/mol, and the molar mass of Cu(OH)₂ is 97.561 g/mol.

Calculation Steps

1. Calculate the moles of CuSO₄: - Moles = Mass / Molar mass - Moles = 24g / 159.609 g/mol - Moles ≈ 0.1505 mol

2. Determine the limiting reactant: - The balanced chemical equation shows that 1 mole of CuSO₄ reacts with 2 moles of NaOH. - Therefore, the moles of NaOH required = 0.1505 mol * 2 = 0.301 mol

3. Calculate the mass of Cu(OH)₂ formed: - Mass = Moles * Molar mass - Mass = 0.1505 mol * 97.561 g/mol - Mass ≈ 14.66g

Approximately 14.66 grams of Cu(OH)₂ precipitate will be formed when 24g of CuSO₄ reacts with the necessary amount of NaOH.

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