Question

Рассчитать массу осадка, полученного при сливании растворов сульфата меди (2) и 17,1 г гидроксида бария. Срочно. Заранее спасибо. :)

Answer 1

CuSO4+Ba(OH)2=Cu(OH)2+BaSO4

n(Ba(OH)2)=17.1/171=0,1 моль
M(Ba(OH)2)=171 г\моль
n(BaSO4)=n(Ba(OH)2)=0.1 моль
m(BaSO4)=0.1*233=23.3 г
M(BaSO4)=233 г\моль
n(Cu(OH)2)=n(Ba(OH)2)=0.1 моль
m(Cu(OH)2)=0.1*98=9,8 г
M(Cu(OH)2)=98 г\моль

m(осадок)=m(Cu(OH)2)+m(BaSO4)=9,8 г+23,3 г=33,1 г

Ответ:m(осадок)=33,1 г

Answer 2

Calculation of Precipitate Mass

To calculate the mass of the precipitate obtained from the reaction between copper sulfate (CuSO4) and barium hydroxide (Ba(OH)2), we need to determine the balanced chemical equation for the reaction. Unfortunately, the search results did not provide the specific balanced equation for this reaction. However, we can still proceed with the calculation by assuming a balanced equation based on the reactants provided.

Let's assume the balanced equation for the reaction is:

CuSO4 + Ba(OH)2 -> Cu(OH)2 + BaSO4

According to the equation, 1 mole of copper sulfate (CuSO4) reacts with 1 mole of barium hydroxide (Ba(OH)2) to produce 1 mole of copper hydroxide (Cu(OH)2) and 1 mole of barium sulfate (BaSO4).

To calculate the mass of the precipitate (BaSO4), we need to know the molar masses of the reactants and products involved. The molar masses are as follows:

- CuSO4: 159.61 g/mol - Ba(OH)2: 171.34 g/mol - Cu(OH)2: 97.56 g/mol - BaSO4: 233.39 g/mol

Now, let's calculate the mass of the precipitate using the given mass of barium hydroxide (Ba(OH)2), which is 17.1 g.

First, we need to determine the number of moles of Ba(OH)2:

Number of moles of Ba(OH)2 = Mass of Ba(OH)2 / Molar mass of Ba(OH)2

Number of moles of Ba(OH)2 = 17.1 g / 171.34 g/mol = 0.0999 mol (approximately)

According to the balanced equation, the molar ratio between Ba(OH)2 and BaSO4 is 1:1. Therefore, the number of moles of BaSO4 formed will also be approximately 0.0999 mol.

Finally, we can calculate the mass of BaSO4:

Mass of BaSO4 = Number of moles of BaSO4 * Molar mass of BaSO4

Mass of BaSO4 = 0.0999 mol * 233.39 g/mol = 23.32 g (approximately)

Therefore, the mass of the precipitate obtained from the reaction between copper sulfate (CuSO4) and barium hydroxide (Ba(OH)2) is approximately 23.32 grams.

Please note that the assumed balanced equation may not be accurate without further information. It is always recommended to consult reliable sources or perform experiments to obtain precise and accurate results.