Question

Consider the reaction of 30.0 mL of 0.246 M BaI₂ with 20.0 mL of 0.315 M Na₃PO₄.

3 BaI₂(aq) + 2 Na₃PO₄(aq) → Ba₃(PO₄)₂(s) + 6 NaI(aq)
If you have 0.00630 moles of Na₃PO₄, What quantity in moles of precipitate are produced if all the Na₃PO₄ were consumed based on the balanced chemical equation?

Answer 1

Using the molar ratio from the balanced equation, 0.00315 moles of Barium Phosphate (Ba3(PO4)2) precipitate are produced when 0.00630 moles of Sodium Phosphate (Na3PO4) are completely reacted.

Step-by-step explanation:

We are given a stoichiometric problem where we need to determine the amount of precipitate formed when a certain amount of Na3PO4 reacts completely. The balanced chemical equation for the reaction provided is:

3 BaI2(aq) + 2 Na3PO4(aq) → Ba3(PO4)2(s) + 6 NaI(aq)

This shows that 2 moles of Na3PO4 react to produce 1 mole of Ba3(PO4)2 precipitate. Given that there are 0.00630 moles of Na3PO4, using the molar ratio from the balanced reaction (3:2), we find the moles of precipitate produced.

Step 1: Identify the molar ratio from the balanced equation for Na3PO4 to Ba3(PO4)2: 2:1.

Step 2: Calculate moles of precipitate using the ratio: 0.00630 moles Na3PO4 x (1 mole Ba3(PO4)2 / 2 moles Na3PO4) = 0.00315 moles of Ba3(PO4)2.

Therefore, 0.00315 moles of Ba3(PO4)2 precipitate are produced when all the Na3PO4 is consumed.