URGENT CHEMISTRY EXPERT!
Consider the reaction :2Na3PO4(aq) + 3CuCl2(aq) → Cu3(PO4)2(s) + 6NaCl(aq)

PART 1: What volume (in mL) of 0.300 MNa3PO4 solutions is needed to completely react with 16.7mLof 0.200M CuCl2?

PART 2: Find the net ionic equation for this reaction.

Part 1: 7.42 mL; Part 2: 3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ 2Cu₃(PO₄)₂(s)

Explanation:

Part 1. Volume of reactant

(a) Balanced chemical equation.

(b) Moles of CuCl₂

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

(d) Volume of Na₃PO₄

Part 2. Net ionic equation

(a) Molecular equation

(b) Ionic equation

You write molecular formulas for the solids, and you write the soluble ionic substances as ions.

According to the solubility rules, metal phosphates are insoluble.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)  

(c) Net ionic equation

To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Cu²⁺(aq) + 6Cl⁻(aq) ⟶ Cu₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)  

The net ionic equation is

3Cu²⁺(aq) + 2PO₄³⁻(aq) ⟶ Cu₃(PO₄)₂(s)

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