Consider the following chemical reaction. 3 Zn(C₂H₂O₂)2(aq) + 2 Na3PO4(aq) →→6 NaC₂H3O₂(aq) + Zn3(PO4)2(S) - Calculate the mass in grams Zn3(PO4)2 that can be precipitated from 0.105 a 1.06 M Zn(C₂H3O2)2 solution by the reaction with excess aqueous sodium phosphate (Na3PO4) according to the following balanced chemical equation: NaC₂H3O₂(aq) + Zn3(PO4)2(S) 3 Zn(C₂H3O₂)2(aq) + 2 Na3PO4(aq) →→ 6 Lof

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**Chemical Reaction Calculation**

Consider the following chemical reaction:

3 Zn(C₂H₃O₂)₂(aq) + 2 Na₃PO₄(aq) → 6 NaC₂H₃O₂(aq) + Zn₃(PO₄)₂(s)

Calculate the mass in grams of Zn₃(PO₄)₂ that can be precipitated from 0.105 L of a 1.06 M Zn(C₂H₃O₂)₂ solution by the reaction with excess aqueous sodium phosphate (Na₃PO₄) according to the balanced chemical equation above.

- **Balanced Equation:**
  - 3 Zn(C₂H₃O₂)₂(aq) + 2 Na₃PO₄(aq) → 6 NaC₂H₃O₂(aq) + Zn₃(PO₄)₂(s)

**Calculation Steps:**

1. **Starting Amount:**
   - Volume (L) of Zn(C₂H₃O₂)₂: 0.105 L
   - Molarity (M) of Zn(C₂H₃O₂)₂: 1.06 M
   
2. **Molar Ratios & Conversion:**
   - Moles of Zn(C₂H₃O₂)₂
   - Moles of Zn₃(PO₄)₂
   - Molar mass of Zn₃(PO₄)₂: 386.1 g/mol

3. **Factors Used:**
   - 2, 0.105, 1.06, 6, 386.1, 3, 14.4, 1

4. **Result:**
   - Mass of Zn₃(PO₄)₂ precipitated: 0.0698 g 

**Visual Aid:**

The calculation involves multiplying the starting amount by conversion factors and molar ratios to determine the final mass in grams. 

This example demonstrates stoichiometry in chemical reactions, emphasizing the importance of balanced equations in calculating product yields.
Transcribed Image Text:**Chemical Reaction Calculation** Consider the following chemical reaction: 3 Zn(C₂H₃O₂)₂(aq) + 2 Na₃PO₄(aq) → 6 NaC₂H₃O₂(aq) + Zn₃(PO₄)₂(s) Calculate the mass in grams of Zn₃(PO₄)₂ that can be precipitated from 0.105 L of a 1.06 M Zn(C₂H₃O₂)₂ solution by the reaction with excess aqueous sodium phosphate (Na₃PO₄) according to the balanced chemical equation above. - **Balanced Equation:** - 3 Zn(C₂H₃O₂)₂(aq) + 2 Na₃PO₄(aq) → 6 NaC₂H₃O₂(aq) + Zn₃(PO₄)₂(s) **Calculation Steps:** 1. **Starting Amount:** - Volume (L) of Zn(C₂H₃O₂)₂: 0.105 L - Molarity (M) of Zn(C₂H₃O₂)₂: 1.06 M 2. **Molar Ratios & Conversion:** - Moles of Zn(C₂H₃O₂)₂ - Moles of Zn₃(PO₄)₂ - Molar mass of Zn₃(PO₄)₂: 386.1 g/mol 3. **Factors Used:** - 2, 0.105, 1.06, 6, 386.1, 3, 14.4, 1 4. **Result:** - Mass of Zn₃(PO₄)₂ precipitated: 0.0698 g **Visual Aid:** The calculation involves multiplying the starting amount by conversion factors and molar ratios to determine the final mass in grams. This example demonstrates stoichiometry in chemical reactions, emphasizing the importance of balanced equations in calculating product yields.
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