Question

Определить массу осадка, который образуется при сливании растворов, содержащих 20,8г барий хлорила и 22,8г сульфатной кислоты

Answer 1

Дано
m(BaCl2) =20.8g
m(H2SO4) =22.8g
найти m(BaSO4)-?
BaCl2+H2SO4 ->2HCl +BaSO4
находим молекулярные массы веществ
Mr(BaCl2)=137+35.5*2=208g/mol
Mr(H2SO4)= 2+32+64=98g/mol
Mr(BaSO4)= 137+32+64=233g/mol
решение
20.8/208=х/233
х=23.3

Answer 2

Calculating the Precipitate Mass

To determine the mass of the precipitate formed when solutions containing 20.8g of barium chloride and 22.8g of sulfuric acid are mixed, we can use the concept of chemical reactions and stoichiometry.

The balanced chemical equation for the reaction between barium chloride (BaCl2) and sulfuric acid (H2SO4) is as follows: ``` BaCl2 + H2SO4 → BaSO4 + 2HCl ```

From the balanced chemical equation, we can see that 1 mole of barium chloride reacts with 1 mole of sulfuric acid to produce 1 mole of barium sulfate and 2 moles of hydrochloric acid.

Using the molar masses of the compounds involved: - Molar mass of BaCl2 = 137.33 g/mol - Molar mass of H2SO4 = 98.08 g/mol - Molar mass of BaSO4 = 233.39 g/mol

We can calculate the amount of barium sulfate formed and then determine the mass of the precipitate.

Calculation Steps

1. Calculate the moles of barium chloride and sulfuric acid. 2. Determine the limiting reactant. 3. Calculate the moles of barium sulfate formed. 4. Convert moles of barium sulfate to grams to find the mass of the precipitate.

Calculation

1. Moles of barium chloride: ``` Moles = Mass / Molar mass Moles = 20.8g / 137.33 g/mol Moles ≈ 0.1516 mol ```

Moles of sulfuric acid: ``` Moles = Mass / Molar mass Moles = 22.8g / 98.08 g/mol Moles ≈ 0.2323 mol ```

2. Determine the limiting reactant: The limiting reactant is the one that is completely consumed in the reaction. To find the limiting reactant, we compare the moles of each reactant to the stoichiometry of the balanced chemical equation.

Since the stoichiometry of the balanced chemical equation is 1:1 for both reactants, the limiting reactant is the one that forms the least amount of product. In this case, barium chloride is the limiting reactant.

3. Moles of barium sulfate formed: From the balanced chemical equation, 1 mole of barium chloride forms 1 mole of barium sulfate. ``` Moles of BaSO4 = Moles of BaCl2 Moles of BaSO4 ≈ 0.1516 mol ```

4. Mass of the precipitate (barium sulfate): ``` Mass = Moles × Molar mass Mass = 0.1516 mol × 233.39 g/mol Mass ≈ 35.39g ```

Result:

The mass of the precipitate (barium sulfate) formed when solutions containing 20.8g of barium chloride and 22.8g of sulfuric acid are mixed is approximately 35.39 grams.