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?
🤓 Based on our data, we think this question is relevant for Professor Schurmeier's class at UCSD.
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- = BaSO4 ; Na+ and Cl- = NaCl
- Balanced reaction will appear as:
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?