Question

Into a 0.25 M solution of Ba3(PO4)2(aq), excess Na2SO4(aq) was added to form BaSO4(s). Ba3(PO4)2(aq) + 3Na2SO4(aq) → 3BaSO4(s) + 2Na3PO4(aq) If you knew the volume of the solution containing Ba3(PO4)2(aq), determine how you would predict the mass of BaSO4(s) formed by completing the following solution map.

Answer 1

Answer is in the explanation.

Step-by-step explanation:

For the reaction:

Ba₃(PO₄)₂(aq) + 3Na₂SO₄(aq) → 3BaSO₄(s) + 2Na₃PO₄(aq)

As Na₂SO₄(aq) is in excess, limiting reactant is Ba₃(PO₄)₂(aq). As the molarity of the solution is 0,25M and you knew the volume of the solution, you can obtain the moles of Ba₃(PO₄)₂ doing 0,25M×volume.

As 1 mol of Ba₃(PO₄)₂(aq) react with 3 moles of BaSO₄ the moles of BaSO₄ are three times moles of Ba₃(PO₄)₂.

As BaSO₄ molar mass is 233,38g/mol. The mass of BaSO₄ is given by moles of BaSO₄ × 233,38g/mol

I hope it helps!