Question

Определите массу осадка которая будет получаться при взаимодействии раствора нитрата бария с раствором сульфата калия массой 348г и массой вещества в нем 20%

Answer 1

Нашла ответ в инете.

Answer 2

Calculation of Precipitate Mass

To determine the mass of the precipitate formed when a solution of barium nitrate reacts with a solution of potassium sulfate, we need to consider the balanced chemical equation for the reaction. Unfortunately, the search results provided do not contain the necessary information to directly calculate the mass of the precipitate. However, I can provide you with the balanced chemical equation and the general approach to solving this type of problem.

The balanced chemical equation for the reaction between barium nitrate (Ba(NO3)2) and potassium sulfate (K2SO4) is as follows:

Ba(NO3)2 + K2SO4 → BaSO4 + 2KNO3

From the balanced equation, we can see that one mole of barium nitrate reacts with one mole of potassium sulfate to produce one mole of barium sulfate and two moles of potassium nitrate.

To calculate the mass of the precipitate, we need to know 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 determine the limiting reactant, we compare the number of moles of each reactant to the stoichiometric coefficients in the balanced equation.

Given that the mass of the potassium sulfate solution is 348 g and it contains 20% of the substance, we can calculate the mass of potassium sulfate in the solution:

Mass of potassium sulfate = 348 g × 0.20 = 69.6 g

To find the number of moles of potassium sulfate, we divide the mass by its molar mass. The molar mass of K2SO4 is 174.26 g/mol:

Number of moles of potassium sulfate = 69.6 g / 174.26 g/mol = 0.399 mol

Since the stoichiometric coefficient of potassium sulfate in the balanced equation is 1, we can conclude that 0.399 moles of potassium sulfate will react.

According to the balanced equation, the stoichiometric coefficient of barium sulfate is also 1. Therefore, the number of moles of barium sulfate formed will also be 0.399 moles.

To calculate the mass of the precipitate (barium sulfate), we multiply the number of moles by its molar mass. The molar mass of BaSO4 is 233.39 g/mol:

Mass of barium sulfate = 0.399 mol × 233.39 g/mol = 93.2 g

Therefore, the mass of the precipitate (barium sulfate) that will be formed is approximately 93.2 grams.

Please note that this calculation assumes that the reaction goes to completion and that there are no other limiting factors or side reactions.