Consider the reaction of 30.0 mL of 0.244 M Bal, with 20.0 mL of 0.315 M Na3PO4. Since the precipitate is Bas(POA)2, write the molecular equation for the precipitation reaction. If no reaction occurs, write NR after the reaction arrow. 3 Bal;(aq) + 2 Na,PO4(aq) → Ba3 (PO4) 2(s) + 6 Nal2 (aq)
Consider the reaction of 30.0 mL of 0.244 M Bal, with 20.0 mL of 0.315 M Na3PO4. Since the precipitate is Bas(POA)2, write the molecular equation for the precipitation reaction. If no reaction occurs, write NR after the reaction arrow. 3 Bal;(aq) + 2 Na,PO4(aq) → Ba3 (PO4) 2(s) + 6 Nal2 (aq)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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What is the molecular equation for This problem ?
![**Question 1.b of 1**
Consider the reaction of 30.0 mL of 0.244 M BaI₂ with 20.0 mL of 0.315 M Na₃PO₄.
Since the precipitate is Ba₃(PO₄)₂, write the molecular equation for the precipitation reaction. If no reaction occurs, write NR after the reaction arrow.
\[ 3 \text{BaI}_{2(aq)} + 2 \text{Na}_3\text{PO}_{4(aq)} \rightarrow \text{Ba}_3(\text{PO}_4)_{2(s)} + 6 \text{NaI}_{(aq)} \]
**Explanation of the Process:**
In this problem, you are given two aqueous solutions: one containing barium iodide (BaI₂) and the other containing sodium phosphate (Na₃PO₄). Upon mixing, these solutions undergo a double displacement reaction, resulting in the formation of barium phosphate (Ba₃(PO₄)₂) as a solid precipitate, and sodium iodide (NaI) remains in solution. The balanced equation demonstrates the stoichiometry of the reactants and products.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F163dc8ba-851f-4764-b6aa-51728fdb757c%2F81edba70-8b0f-4390-892b-381d4079f5b8%2F2uv60lh_processed.png&w=3840&q=75)
Transcribed Image Text:**Question 1.b of 1**
Consider the reaction of 30.0 mL of 0.244 M BaI₂ with 20.0 mL of 0.315 M Na₃PO₄.
Since the precipitate is Ba₃(PO₄)₂, write the molecular equation for the precipitation reaction. If no reaction occurs, write NR after the reaction arrow.
\[ 3 \text{BaI}_{2(aq)} + 2 \text{Na}_3\text{PO}_{4(aq)} \rightarrow \text{Ba}_3(\text{PO}_4)_{2(s)} + 6 \text{NaI}_{(aq)} \]
**Explanation of the Process:**
In this problem, you are given two aqueous solutions: one containing barium iodide (BaI₂) and the other containing sodium phosphate (Na₃PO₄). Upon mixing, these solutions undergo a double displacement reaction, resulting in the formation of barium phosphate (Ba₃(PO₄)₂) as a solid precipitate, and sodium iodide (NaI) remains in solution. The balanced equation demonstrates the stoichiometry of the reactants and products.
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