Question

Сколько граммов гидроксида меди(2) можно получить 80г 30- процентного сульфата меди(2)

Answer 1

CuSO4 + 2NaOH --> Cu(OH)2 + Na2SO4
M(CuSO4)= 64+32+16*4= 160
M((Cu(OH)2)= 64+ 2(16+1)= 98
m(CuSO4)=80*0.3= 24 г
n(CuSO4)=24/160= 0.15 моль
n((Cu(OH)2)=n(CuSO4)=0.15 моль
m((Cu(OH)2)= 0.15*98= 14.7 г

Answer 2

Calculation of the Grams of Copper(II) Hydroxide

To calculate the grams of copper(II) hydroxide that can be obtained from 80g of 30% copper(II) sulfate, we need to use the molar ratios between the two compounds.

The molar mass of copper(II) sulfate (CuSO4) is 159.61 g/mol, and the molar mass of copper(II) hydroxide (Cu(OH)2) is 97.56 g/mol.

First, let's calculate the number of moles of copper(II) sulfate: ``` moles of CuSO4 = (mass of CuSO4) / (molar mass of CuSO4) = (80g) / (159.61 g/mol) ≈ 0.501 moles ```

Next, we need to determine the molar ratio between copper(II) sulfate and copper(II) hydroxide. From the balanced chemical equation, we know that 1 mole of copper(II) sulfate reacts with 1 mole of copper(II) hydroxide: ``` CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4 ```

Therefore, the number of moles of copper(II) hydroxide formed will be the same as the number of moles of copper(II) sulfate: ``` moles of Cu(OH)2 = moles of CuSO4 ≈ 0.501 moles ```

Finally, we can calculate the grams of copper(II) hydroxide: ``` grams of Cu(OH)2 = (moles of Cu(OH)2) * (molar mass of Cu(OH)2) = (0.501 moles) * (97.56 g/mol) ≈ 48.84 grams ```

Therefore, approximately 48.84 grams of copper(II) hydroxide can be obtained from 80 grams of 30% copper(II) sulfate.

Please note that this calculation assumes ideal conditions and complete conversion of reactants to products. In practice, the actual yield may be lower due to factors such as incomplete reactions or side reactions.