Given:
- Volume of Na2SO4 solution = 500 mL
- Concentration of Na2SO4 solution = 0.250 M
- Mass of BaCl2 = 15.00 g

Required:
- Mass of BaSO4 formed

Solution:
Step 1: Calculate the number of moles of Na2SO4
- The volume of Na2SO4 solution is given as 500 mL, which is equivalent to 0.5 L.
- The concentration of Na2SO4 solution is given as 0.250 M.
- Therefore, the number of moles of Na2SO4 can be calculated using the formula:
Moles = Concentration × Volume
Moles = 0.250 mol/L × 0.5 L
Moles = 0.125 mol

Step 2: Calculate the number of moles of BaCl2
- The molar mass of BaCl2 can be calculated by adding the atomic masses of Barium (Ba) and Chlorine (Cl).
Molar mass of BaCl2 = (1 × Atomic mass of Ba) + (2 × Atomic mass of Cl)
Molar mass of BaCl2 = (1 × 137.33 g/mol) + (2 × 35.45 g/mol)
Molar mass of BaCl2 = 137.33 g/mol + 70.90 g/mol
Molar mass of BaCl2 = 208.23 g/mol
- The mass of BaCl2 is given as 15.00 g.
- Therefore, the number of moles of BaCl2 can be calculated using the formula:
Moles = Mass / Molar mass
Moles = 15.00 g / 208.23 g/mol
Moles = 0.072 moles

Step 3: Determine the limiting reactant
- To find the limiting reactant, we compare the moles of Na2SO4 and BaCl2.
- The balanced chemical equation for the reaction between Na2SO4 and BaCl2 is as follows:
Na2SO4 + BaCl2 → BaSO4 + 2 NaCl
- From the balanced equation, we can see that the stoichiometric ratio between Na2SO4 and BaSO4 is 1:1.
- Therefore, the limiting reactant will be the one with a smaller number of moles.
- In this case, Na2SO4 has 0.125 moles and BaCl2 has 0.072 moles.
- Hence, BaCl2 is the limiting reactant.

Step 4: Calculate the moles and mass of BaSO4 formed
- From the balanced equation, we can see that 1 mole of BaCl2 reacts to form 1 mole of BaSO4.
- Therefore, the number of moles of BaSO4 formed will be equal to the moles of BaCl2, which is 0.072 moles.
- The molar mass of BaSO4 can be calculated by adding the atomic masses of Barium (Ba), Sulfur (S), and Oxygen (O).
Molar mass of BaSO4 = (1 × Atomic mass