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?

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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?