BaCl2(aq) +Na2SO4(aq) -------> BaSO4(s) + 2NaCl(aq). Balance the reaction. A 0.3

Question

BaCl2(aq) +Na2SO4(aq) -------> BaSO4(s) + 2NaCl(aq). Balance the reaction. A 0.30 M solution is added to 25.0 mL of 0.50 M Na2SO4 solution until no more precipitate forms.

Part A: WHat mass (in grams) of BaSO4 is produced?

Part B: What volume of BaCl2 solution was added?

Please explain the steps clearly. Thank you.

Explanation / Answer

balanced equation is BaCl2(aq) +Na2SO4(aq) -------> BaSO4(s) + 2NaCl(aq) moles of Na2SO4 in 25.0 mL of 0.50 M Na2SO4 solution = 0.5 x 0.025 =0.0125 moles 1 mole of Na2SO4 gives 1 mole of BaSO4 mass (in grams) of BaSO4 produced= mass of 1 mole of BaSO4= 233.39 g Part A: so 0.0125 moles of Na2SO4 will produce 0.0125 x 233.39 =2.917 g of BaSO4 Part B: moles of BaCl2 reacted = moles of Na2SO4 in 25.0 mL of 0.50 M Na2SO4 solution = 0.0125 moles so volume of 0.30 M solution of BaCl2 contaning 0.0125 moles = 41.6 ml

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