Question

Вычислите массу осадка образовавшегося при взаимодействии 50г хлорида бария и раствора сульфата натрия.

Answer 1

BaCl₂ + Na₂SO₄ = BaSO₄↓ + 2NaCl

M (BaCl₂) = 208 г/моль
M (BaSO₄) = 233 г/моль

208 г (BaCl₂) → 233 г (BaSO₄)
50 г (BaCl₂) → m (BaSO₄)

m (BaSO₄) = 50·233/208 ≈ 56 г

Answer 2

Calculating the Precipitate Mass

To calculate the mass of the precipitate formed during the interaction of 50g of barium chloride and a solution of sodium sulfate, we can use the principles of stoichiometry and chemical reactions.

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the reaction between barium chloride (BaCl2) and sodium sulfate (Na2SO4) is:

BaCl2 + Na2SO4 → BaSO4 + 2NaCl

Step 2: Determine the Limiting Reagent

To determine the limiting reagent, we need to compare the moles of each reactant to the stoichiometry of the balanced chemical equation.

From the given information, we know that the mass of barium chloride (BaCl2) is 50g.

Step 3: Calculate the Mass of Precipitate

Using the balanced chemical equation and the stoichiometry, we can calculate the mass of the precipitate (barium sulfate, BaSO4) formed.

The molar mass of BaCl2 is 137.327 g/mol, and the molar mass of BaSO4 is 233.389 g/mol.

Now, let's calculate the mass of the precipitate formed.

First, we need to find the moles of barium chloride: ``` moles of BaCl2 = mass of BaCl2 / molar mass of BaCl2 = 50g / 137.327 g/mol ≈ 0.364 moles ```

According to the stoichiometry of the balanced chemical equation, 1 mole of BaCl2 reacts with 1 mole of Na2SO4 to produce 1 mole of BaSO4.

Therefore, the moles of BaSO4 formed will also be approximately 0.364 moles.

Finally, we can calculate the mass of the precipitate (BaSO4): ``` mass of BaSO4 = moles of BaSO4 * molar mass of BaSO4 ≈ 0.364 moles * 233.389 g/mol ≈ 84.99g ```

Conclusion: The mass of the precipitate formed during the interaction of 50g of barium chloride and a solution of sodium sulfate is approximately 84.99 grams.