Question

Определите массу и кол-во вещества хлорида бария для получения 2,5 моль осадка сульфата бария

Answer 1

Дано:
ню(BaSO4)=2.5 моль
Найти
m(BaCl2)-?
ню(BaCl2)-?
Уравнение реакции:BaCl2+H2SO4=BaSO4+2HCl
1)ню(BaCl2)=ню(BaSO4)=2,5 моль т.к коэффиценты одинаковые
ню(BaCl2)=2,5 моль
2)m(BaCl2)=ню(BaCl2)*M(BaCl2)
M(BaCl2)=137+35.5*2=137+71=208г/моль
M(BaCl2)=208 г/моль
m(BaCl2)=2.5моль*208г/моль=520г
m(BaCl2)=520г
Ответ:m(BaCl2)=520г;ню(BaCl2)=2.5 моль

Answer 2

Calculating the Mass and Amount of Barium Chloride for Precipitating 2.5 Moles of Barium Sulfate

To determine the mass and amount of substance of barium chloride needed to produce 2.5 moles of barium sulfate precipitate, we can use the balanced chemical equation for the reaction between barium chloride and sodium sulfate:

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. Therefore, to produce 2.5 moles of barium sulfate, we would need 2.5 moles of barium chloride.

Calculating the Mass of Barium Chloride

The molar mass of barium chloride (BaCl2) is 137.327 + 2*35.453 = 208.233 g/mol. Therefore, the mass of 2.5 moles of barium chloride can be calculated as:

Mass = 2.5 moles * molar mass of BaCl2 Mass = 2.5 * 208.233 g/mol Mass ≈ 520.5825 g

Conclusion

To produce 2.5 moles of barium sulfate precipitate, approximately 520.5825 grams of barium chloride would be required.