Question

Consider the following precipitation reaction:

2Na₃PO₄(aq)+3CuCl2(aq)→Cu₃(PO₄)2(s)+6NaCl(aq)
What volume of 0.182 MNa₃PO₄ solution is necessary to completely react with 89.0 mL of 0.120 MCuCl₂?

Answer 1

To react with 89.0 mL of 0.120 M CuCl2, 39.12 mL of 0.182 M Na3PO4 is required.

Step-by-step explanation:

To determine what volume of 0.182 M Na3PO4 solution is necessary to completely react with 89.0 mL of 0.120 M CuCl2, start by writing out the balanced chemical equation. The chemical equation for the reaction between sodium phosphate (Na3PO4) and copper(II) chloride (CuCl2) would likely be:

3 CuCl2(aq) + 2 Na3PO4(aq) → Cu3(PO4)2(s) + 6 NaCl(aq)

Next, use stoichiometry to find the moles of CuCl2 reacted:

(0.120 mol/L) × (0.089 L) = 0.01068 mol of CuCl2

From the balanced equation, 3 moles of CuCl2 react with 2 moles of Na3PO4, so the moles of Na3PO4 needed are:

(2/3) × 0.01068 mol = 0.00712 mol of Na3PO4

Finally, calculate the volume of Na3PO4 solution needed:

Volume =
moles / concentration

Volume = 0.00712 mol / 0.182 M = 0.03912 L or 39.12 mL