Question

Вычислите массу осадка, выпавшего при взаимодействии 980г 20%-го раствора сульфата меди (//) с необходимым количеством гидроксида калия. Какого количество вещества образовавшегося осадка?

Answer 1

CuSO4+2KOH=K2SO4+Cu(OH)2
m(CuSO4чист)=980*0.2=196гр
n(CuSO4)=196/159=1.2моль
n(Cu(OH)2)=1.2моль
m(Cu(OH)2)=1.2*97=116.4гр

Answer 2

Calculating the Mass of Precipitate and the Amount of Substance Formed

To calculate the mass of the precipitate formed during the interaction of a 980g 20% solution of copper sulfate with the necessary amount of potassium hydroxide, we can use the stoichiometry of the reaction. The balanced chemical equation for the reaction between copper sulfate and potassium hydroxide is:

CuSO4 + 2KOH → Cu(OH)2 + K2SO4

From the balanced equation, we can see that 1 mole of copper sulfate reacts with 2 moles of potassium hydroxide to produce 1 mole of copper hydroxide and 1 mole of potassium sulfate.

First, let's calculate the mass of copper sulfate in the 980g 20% solution:

Mass of copper sulfate = 980g * 20% = 196g

Now, we need to calculate the amount of substance formed, which is the amount of copper hydroxide precipitate. Since the molar mass of copper hydroxide (Cu(OH)2) is approximately 97.56 g/mol, we can calculate the amount of substance formed using the following formula:

Amount of substance = Mass / Molar mass

Amount of substance of copper hydroxide = 196g / 97.56 g/mol = 2.01 mol

Therefore, the amount of substance of copper hydroxide formed is approximately 2.01 moles.

This calculation assumes that the reaction goes to completion and that all the copper sulfate reacts with the potassium hydroxide to form copper hydroxide.

I hope this helps! If you have further questions or need additional assistance, feel free to ask.