Molarity/molality help #5

Question

How many milliliters of 0.736 M H₃PO₄ are required to react with 239 mL of 0.157 M Ba(OH)₂ if the products are barium phosphate and water?

__ mL

Answer 1

We'll need to write a correctly balanced equation for the reaction taking place:

2H₃PO₄ + 3Ba(OH)₂ ==> Ba₃(PO₄)₂ + 6H₂O ... balanced equation

NOTE: TWO moles H₃PO₄ reacts with THREE moles Ba(OH)₂

moles Ba(OH)₂ present = 239 mls x 1 L / 1000 mls x 0.157 mols / L = 0.0375 mols Ba(OH)₂

moles H₃O₄ needed = 0.0375 mols Ba(OH)₂ x 2 mol H₃PO₄ / 3 mols Ba(OH)₂ = 0.0250 mols H₃PO₄

mls H₃PO₄ needed = 0.0250 mols x 1 L / 0.736 mols x 1000 mls / L = 34.0 mls H₃PO₄