Question

Вычислить массу осадка образовавшегося при взаимодействии раствора хлорида бария со 120г раствора сульфата натрия

Answer 1

                                     120 г                 Х г
     BaCl2          +         Na2SO4     ->   BaSO4    +   2NaCl
                                n=1 моль               n=1 моль
                              М = 142 г/моль       М = 233 г/моль
                              m=142 г                   m=233 г

120 г Na2SO4     -     Х г BaSO4
142 г Na2SO4     -     233 г BaSO4
m(BaSO4) = 120 * 233 / 142 = 197 г

Answer 2

Calculating the Precipitate Mass

To calculate the mass of the precipitate formed when a solution of barium chloride reacts with a 120g solution of sodium sulfate, we can use the principles of stoichiometry and the concept of limiting reagents.

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

BaCl2 + Na2SO4 → BaSO4 + 2NaCl

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

Given the molar masses: - BaCl2 = 137.33 g/mol - Na2SO4 = 142.04 g/mol - 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 sodium sulfate in the 120g solution using its molar mass. 2. Use the mole ratio from the balanced equation to find the moles of barium sulfate formed. 3. Convert the moles of barium sulfate to grams to find the mass of the precipitate.

Calculation

1. Moles of Na2SO4: - Moles = Mass / Molar mass - Moles = 120g / 142.04 g/mol - Moles ≈ 0.845 moles 2. Moles of BaSO4 formed: - From the balanced equation, 1 mole of Na2SO4 reacts with 1 mole of BaSO4. - Moles of BaSO4 = Moles of Na2SO4 - Moles of BaSO4 ≈ 0.845 moles

3. Mass of BaSO4 formed: - Mass = Moles × Molar mass - Mass ≈ 0.845 moles × 233.39 g/mol - Mass ≈ 197.5g

The mass of the precipitate formed is approximately 197.5 grams.