Question

определите количество вещества гидроксида меди 2 образующегося при взаимодействии хлорида меди 2 и 10 г. раствора гидроксида натрия с массовой долей 8%

Answer 1

CuCl2+2NaOH=Cu(OH)2+2NaCl

1)Найдем массу NaOH в растворе:

m(NaOH)=ω(р-ра)*m(р-ра)=0,08*10г=0,8г

2)Найдем кол-во NaOH:

v(NaOH)=m(NaOH)/M(NaOH)=0,8/40г/моль=0,02 моль

Answer 2

m(NaOH)=0,08 * 10=0,8 грамм

n(NaOH)=0,8/40=0,02 моль

CuCl2 + 2NaOH --> Cu(OH)2 + 2NaCl

0,02/2=y/1

y=0,01 моль

Ответ: n(Cu(OH)2)=0,01 моль

Answer 3

Calculating the Amount of Copper(II) Hydroxide Formed

To calculate the amount of substance of copper(II) hydroxide formed when copper(II) chloride reacts with a 10 g solution of sodium hydroxide with a mass fraction of 8%, we can use the balanced chemical equation for the reaction and stoichiometry.

The balanced chemical equation for the reaction is: CuCl2 + 2NaOH → Cu(OH)2 + 2NaCl

From the equation, we can see that 1 mole of copper(II) chloride reacts with 2 moles of sodium hydroxide to produce 1 mole of copper(II) hydroxide.

Given: - Mass of sodium hydroxide solution = 10 g - Mass fraction of sodium hydroxide in the solution = 8%

First, we need to calculate the mass of sodium hydroxide in the solution: Mass of NaOH = Mass of solution × Mass fraction of NaOH Mass of NaOH = 10 g × 8% = 10 g × 0.08 = 0.8 g

Now, we can use the molar mass of NaOH to convert the mass of NaOH to moles: Molar mass of NaOH = 22.99 g/mol (Na) + 15.999 g/mol (O) + 1.008 g/mol (H) = 39.997 g/mol ≈ 40 g/mol Moles of NaOH = Mass of NaOH / Molar mass of NaOH = 0.8 g / 40 g/mol = 0.02 moles

Since 1 mole of copper(II) hydroxide is formed for every 2 moles of NaOH, the moles of copper(II) hydroxide formed will be half of the moles of NaOH: Moles of Cu(OH)2 = 0.02 moles / 2 = 0.01 moles

Therefore, the amount of substance of copper(II) hydroxide formed is approximately 0.01 moles.