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 is in the explanation.

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!

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