Question

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.

Answer 1

Answer:

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.

$\rm 2Na_{3}PO_{4} + 3CuCl_{2} \longrightarrow Cu_{3}(PO_{4})_{2} + 6NaCl$

(b) Moles of CuCl₂

$\text{Moles of CuCl}_{2} =\text{ 16.7 mL CuCl}_{2} \times \dfrac{\text{0.200 mmol CCl}_{2}}{\text{1 mL CuCl}_{2}} = \text{3.340 mmol CuCl}_{2}$

(c) Moles of Na₃PO₄

The molar ratio is 2 mmol Na₃PO₄:3 mmol CuCl₂

$\text{Moles of Na _{3} PO}_{4} = \text{3.340 mmol CuCl}_{2} \times \dfrac{\text{2 mmol Na _{3} PO}_{4}}{\text{3 mmol CuCl}_{2}} =\text{2.227 mmol Na _{3} PO}_{4}$

(d) Volume of Na₃PO₄

$V = \text{2.227 mmol Na _{3} PO}_{4}\times \dfrac{\text{1 mL Na _{3} PO}_{4}}{\text{0.300 mmol Na _{3} PO}_{4}} = \text{7.42 mL Na _{3} PO}_{4} \\\\\text{The reaction requires \large \boxed{\textbf{7.42 mL Na _{3} PO}_{4}} }$

Part 2. Net ionic equation

(a) Molecular equation

$\rm 2Na_{3}PO_{4}(\text{aq}) + 3CuCl_{2}(\text{aq}) \longrightarrow Cu_{3}(PO_{4})_{2}(\text{s}) + 6NaCl(\text{aq})$

(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)