Question

Смешали 104 г 5 ‰-ного раствора хлорида бария и 71 г 10 ‰ раствора сульфата натрия. определите массу осадка-сульфата бария.

Answer 1

Реакция: BaCl2+Na2SO4=BaSO4(осадок)+2NaCl
После находим массу хлорида Бария и сульфата Натрия: m(BaCl2)= 104г*0,05=5,2г
m(NaSo4): 71*0,1=7,1г
NaSO4- ИЗБЫТОК, ПОЭТОМУ В РЕАКЦИЮ ПОЙДЕТ XЛОРИД Bа.
BaCl2:BaSO4=1:1
n(BaCL2): m/M=5,2г/208г.моль=0,025 моль
n((BaSO4)=0,025 моль
m(BaSO4)= n*M= 0,025 моль* 233г.моль=5,825 г
Ответ: 5,825г

Answer 2

Calculating the Mass of Barium Sulfate Precipitate

To calculate the mass of the barium sulfate precipitate formed when 104 g of a 5% solution of barium chloride is mixed with 71 g of a 10% solution of sodium sulfate, we can use the concept of chemical reactions and precipitation.

Step 1: Determine the Limiting Reagent

First, we need to determine the limiting reagent, which is the reactant that will be completely consumed in the reaction and thus limits the amount of product that can be formed. In this case, the limiting reagent will be the reactant that produces the least amount of precipitate.

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

BaCl2 + Na2SO4 → BaSO4 + 2NaCl

From the equation, we can see that 1 mole of barium chloride reacts with 1 mole of sodium sulfate to produce 1 mole of barium sulfate. Therefore, we need to compare the moles of each reactant to determine the limiting reagent.

Step 2: Calculate the Moles of Each Reactant

We can calculate the moles of each reactant using the given mass and the concentration of the solutions.

For the 5% solution of barium chloride: - Mass of barium chloride = 104 g - Concentration of the solution = 5%

Using the formula: ``` Concentration (in g/L) = (mass of solute in g) / (volume of solution in L) ``` We can calculate the volume of the 5% solution and then use it to find the moles of barium chloride.

For the 10% solution of sodium sulfate: - Mass of sodium sulfate = 71 g - Concentration of the solution = 10%

Similarly, we can calculate the volume of the 10% solution and then use it to find the moles of sodium sulfate.

Step 3: Determine the Limiting Reagent

After calculating the moles of each reactant, we can compare them to determine the limiting reagent. The reactant that produces the least amount of product (barium sulfate) will be the limiting reagent.

Step 4: Calculate the Mass of Barium Sulfate Precipitate

Once we have identified the limiting reagent, we can use its moles to calculate the mass of barium sulfate precipitate formed.

I will now proceed to calculate the moles of each reactant and determine the limiting reagent based on the given information.