Question

What mass of barium sulfate would be produced from 10 g of barium chloride in the following reaction?

BaCl2 + H2SO4 → BaSO4 + 2HCl

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Answer 1

Answer:

10/208=0.048 moles of barium chloride

10/208=0.048 moles of barium chloridetherefore 0.048 moles of barium sulphate so:

0.048x233=11.2g

Explanation:

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Answer 2

Given: 10gm of Barium Chloride

To find: Mass of Barium Sulfate

Solution:  If Barium Chloride reacts with Sulphuric acid it forms Barium Sulphate and Hydrochloride.

• One mole of barium chloride and one mole of sulphric acid forms one mole of Barium Sulphate and two moles of Hydrogen Chloride.
• The molecular mass of Barium Chloride is 208 grams, and the molecular mass of Barium Sulphate is 233 grams
• 10 grams of barium sulfate is equal to the 10/208 mole = 0.048 moles
• By disproportion method, 0.048 moles of Barium Sulphate react with 0.048 moles of sulphuric acid to form 0.48 moles of Barium Sulphate and 0.048 moles of hydrogen chloride.
• 0.048 moles of Barium Sulphate is equal to the mass of barium sulfate / molecular mass of barium sulfate
• 0.048 = x/233
• Therefore, the mass of Barium Sulfate is equal to 11.2 grams.

Therefore, the mass of barium sulfate would be 11.2 grams when it reacts with 10 grams of barium chloride.