Answer to Question #131619 in General Chemistry for Logan

Question

Consider the reaction of 30.0 mL of 0.235 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, how many moles of precipitate are produced if all the the Na₃PO₄ were consumed based on the balanced chemical equation?

Answer 1

Given:

V(BaI₂) = 30 ml => 0.03 L

V(Na₃PO₄) = 20 ml => 0.02 L

M(BaI₂) = 0.235 M

M(Na₃PO₄) = 0.315 M

Solution:

1-step: We should find moles in terms of molarities

M_(molarity) = $\tfrac{n_{moles}}{V_{solution}}$

n(BaI₂) = 0.00705 moles

n(Na₃PO₄) = 0.0063 moles

2-step: We can find moles of precipitate by using limiting reagent rule

0.00705 0.0063 0.00315

3 BaI₂ (aq) + 2 Na₃PO₄ (aq) $\to$ Ba₃(PO₄)₂ (s) + 6 NaI (aq)

If all sodium phosphate are consumed, it is limiting reactant in this reaction and its coefficient is 2 so we devide it by 2 than we can find moles of precipitate formed

n(Ba₃(PO₄)₂$)=\tfrac{0.0063 moles}{2}= 0.00315 moles$