3- Reaction between barium chloride and sulfuric acid: A volume V= 100 ml of solution S1 with concentration C= 2× 10-3 molL is added to 0.002 mol of sulfuric acid H2SO4, barium sulfate is formed according to the following equation: BaCl2(aq) + H,SO(aq) → 2 HCl (aq) + BaSO4) Given: M (BaS04) = 233 g.mol- 3.1- Specify the limiting reagent. 3.2- Determine the amount (in mol) of excess reactant that remained at the end of reaction. 3.3- Calculate the mass of BaSO4 that should be obtained at the end of the reaction. 3.4-Determine the yield of this reaction if 0.022g of barium sulfate is obtained.

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3- Reaction between barium chloride and sulfuric acid:
A volume V= 100 ml of solution Sį with concentration C= 2× 10-3 mol/Lis added to 0.002
mol of sulfuric acid H2SO4, barium sulfate is formed according to the following equation:
BaCl(aq) + H2SO4(aq)
2 HCI (a9) + BasO46)
Given: M (BaSO4) = 233 g.mol!
3.1- Specify the limiting reagent.
3.2- Determine the amount (in mol) of excess reactant that remained at the end of reaction.
3.3- Calculate the mass of BaSO, that should be obtained at the end of the reaction.
3.4-Determine the yield of this reaction if 0.022g of barium sulfate is obtained.
Transcribed Image Text:3- Reaction between barium chloride and sulfuric acid: A volume V= 100 ml of solution Sį with concentration C= 2× 10-3 mol/Lis added to 0.002 mol of sulfuric acid H2SO4, barium sulfate is formed according to the following equation: BaCl(aq) + H2SO4(aq) 2 HCI (a9) + BasO46) Given: M (BaSO4) = 233 g.mol! 3.1- Specify the limiting reagent. 3.2- Determine the amount (in mol) of excess reactant that remained at the end of reaction. 3.3- Calculate the mass of BaSO, that should be obtained at the end of the reaction. 3.4-Determine the yield of this reaction if 0.022g of barium sulfate is obtained.
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