Problem: How many grams of barium sulfate form when 35.0 mL of 0.160 M barium chloride reacts with 58.0 mL of 0.065 M sodium sulfate?

FREE Expert Solution
  • First step is to write the balanced double displacement reaction to establish a relationship between the reactants and products
  • Figure out which will be the limiting reactant between BaCl2 and Na2SO4. The limiting reactant will produce the least but correct amount of product 
  • Using BaCl2 and Na2SO4's provided molarity and volume, we can find the moles of each reactant and then use the balanced equation to establish a relationship between the reactants and BaSO4 (desired product)
  • Then use the molar mass of barium sulfate to determine the mass produced using the moles from the stoichiometry
  • Barium chloride will appear as BaCl2 (Ba2+ and Cl-) while sodium sulfate as Na2SO4 (Na+ and SO42-)
  • In a double displacement reaction, counterions are switched. Salts formed are:

Ba2+ and SO42- = BaSO; Na+ and Cl- = NaCl


  • Balanced reaction will appear as:
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Problem Details

How many grams of barium sulfate form when 35.0 mL of 0.160 M barium chloride reacts with 58.0 mL of 0.065 M sodium sulfate?

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Solution Stoichiometry concept. You can view video lessons to learn Solution Stoichiometry. Or if you need more Solution Stoichiometry practice, you can also practice Solution Stoichiometry practice problems.

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Based on our data, we think this problem is relevant for Professor Schurmeier's class at UCSD.

What textbook is this problem found in?

Our data indicates that this problem or a close variation was asked in Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition. You can also practice Chemistry: The Molecular Nature of Matter and Change - Silberberg 8th Edition practice problems.