Question

При взаимодействии 160 г 10% раствора гидроксида натрия с сульфатом меди (II) образовался сульфат натрия и гидроксид меди(II). Рассчитайте массу образовавшегося осадка.

Answer 1

Решение:  NaOH+CuSO4=Na2SO4+Cu(OH2)
1) 160 * 10%/ 100%= 16г
2) 16 * 97(гидрок.меди - осадок) / 40 (гидрок.натрия) = 38,8г (гидрок.меди)
Ответ: 38,8г гидроксид меди 2

Answer 2

Calculation of the Mass of Precipitate Formed

To calculate the mass of the precipitate formed when 160 g of a 10% solution of sodium hydroxide reacts with copper(II) sulfate, we need to determine the limiting reactant and use stoichiometry.

The balanced chemical equation for the reaction is:

2NaOH + CuSO4 -> Na2SO4 + Cu(OH)2

From the equation, we can see that the molar ratio between sodium hydroxide (NaOH) and copper(II) sulfate (CuSO4) is 2:1. This means that for every 2 moles of NaOH, 1 mole of CuSO4 is required.

To calculate the mass of the precipitate formed, we can follow these steps:

1. Calculate the number of moles of NaOH: - The concentration of the 10% solution of NaOH is 10 g/100 mL. - Since the solution has a mass of 160 g, we can calculate the mass of NaOH in the solution as 10% of 160 g: (10/100) * 160 g = 16 g. - The molar mass of NaOH is 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.00 g/mol. - Therefore, the number of moles of NaOH is 16 g / 39.00 g/mol = 0.410 moles.

2. Calculate the number of moles of CuSO4: - The molar mass of CuSO4 is 63.55 g/mol + 32.07 g/mol + (4 * 16.00 g/mol) = 159.61 g/mol. - Since the molar ratio between NaOH and CuSO4 is 2:1, the number of moles of CuSO4 is half the number of moles of NaOH: 0.410 moles / 2 = 0.205 moles.

3. Calculate the mass of the precipitate (Cu(OH)2): - The molar mass of Cu(OH)2 is 63.55 g/mol + (2 * 1.01 g/mol) + (2 * 16.00 g/mol) = 97.56 g/mol. - Therefore, the mass of the precipitate formed is 0.205 moles * 97.56 g/mol = 20.00 g.

The mass of the precipitate formed is 20.00 g.

Please note that the calculations provided are based on the information given and the assumption that the reaction goes to completion.