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

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

Step-by-step 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) * \frac{\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) * \frac{\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)* \frac{\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)