Question

Какова масса осадка (в г), который образуется при взаимодействии раствора, содержащего 8 г гидроксида натрия, и раствора, содержащего 24 г сульфата меди(II)? Ответ запишите числом с точностью до десятых. Атомные массы элементов округлите в расчётах до целых.

Answer 1

Ответ:

9,8

Объяснение:

Answer 2

Calculation of the Mass of Precipitate

To calculate the mass of the precipitate formed during the interaction of a solution containing 8 g of sodium hydroxide (NaOH) and a solution containing 24 g of copper sulfate (CuSO4), we need to determine the stoichiometry of the reaction between these two compounds.

The balanced chemical equation for the reaction between sodium hydroxide and copper sulfate is as follows:

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

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

To calculate the mass of the precipitate, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.

To find the limiting reactant, we need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.

The molar mass of sodium hydroxide (NaOH) is approximately 40 g/mol Therefore, 8 g of sodium hydroxide is equal to 0.2 moles (8 g / 40 g/mol).

The molar mass of copper sulfate (CuSO4) is approximately 160 g/mol Therefore, 24 g of copper sulfate is equal to 0.15 moles (24 g / 160 g/mol).

Comparing the number of moles of each reactant to the stoichiometric ratio in the balanced equation, we can see that 0.2 moles of sodium hydroxide react with 0.1 moles of copper sulfate to produce 0.1 moles of copper hydroxide.

Since the stoichiometric ratio is 2:1 for sodium hydroxide to copper sulfate, we can conclude that copper sulfate is the limiting reactant.

Now, we can calculate the mass of the precipitate (copper hydroxide) formed using the molar mass of copper hydroxide, which is approximately 97 g/mol.

0.1 moles of copper hydroxide is equal to 9.7 g (0.1 moles x 97 g/mol).

Therefore, the mass of the precipitate formed is approximately 9.7 grams.

Calculation Steps:

1. Calculate the number of moles of sodium hydroxide: 8 g / 40 g/mol = 0.2 moles. 2. Calculate the number of moles of copper sulfate: 24 g / 160 g/mol = 0.15 moles. 3. Compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation. 4. Determine that copper sulfate is the limiting reactant. 5. Calculate the number of moles of copper hydroxide formed: 0.1 moles (since the stoichiometric ratio is 2:1 for sodium hydroxide to copper sulfate). 6. Calculate the mass of the precipitate (copper hydroxide): 0.1 moles x 97 g/mol = 9.7 grams.

Please note that the atomic masses of the elements used in the calculations have been rounded to whole numbers for simplicity.