onsider the reaction of 30.0 mL of 244 M Bal, with 20.0 mL of 0.315 Na3PO4. Bal2(aq) + 2 NazPO4(aq) → s(POa)2(s) + 6 Nal (aq) you have 0.00630 moles of PO4, What quantity in moles of ecipitate are produced if all the A3PO4 were consumed based on a balanced chemical eauation?
onsider the reaction of 30.0 mL of 244 M Bal, with 20.0 mL of 0.315 Na3PO4. Bal2(aq) + 2 NazPO4(aq) → s(POa)2(s) + 6 Nal (aq) you have 0.00630 moles of PO4, What quantity in moles of ecipitate are produced if all the A3PO4 were consumed based on a balanced chemical eauation?
Chemistry
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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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![**Balanced Chemical Reaction Problem**
**Question:**
Consider the reaction of 30.0 mL of 0.244 M BaI₂ with 20.0 mL of 0.315 M Na₃PO₄.
The balanced chemical equation for the reaction is:
\[ 3 \, \text{BaI}_2(aq) + 2 \, \text{Na}_3\text{PO}_4(aq) \rightarrow \text{Ba}_3(\text{PO}_4)_2(s) + 6 \, \text{NaI}(aq) \]
If you have 0.00630 moles of \(\text{Na}_3\text{PO}_4\), what quantity in moles of precipitate is produced if all the \(\text{Na}_3\text{PO}_4\) were consumed based on the balanced chemical equation?
**Answer:**
To solve this, first determine the moles of precipitate (\(\text{Ba}_3(\text{PO}_4)_2\)) that are formed according to the stoichiometric coefficients in the balanced equation. According to the equation, 2 moles of \(\text{Na}_3\text{PO}_4\) produce 1 mole of \(\text{Ba}_3(\text{PO}_4)_2\).
Given:
- Moles of \(\text{Na}_3\text{PO}_4\) = 0.00630 moles
Using stoichiometry:
\[ \text{Moles of } \text{Ba}_3(\text{PO}_4)_2 = \frac{1}{2} \times \text{Moles of Na}_3\text{PO}_4\]
\[ = \frac{1}{2} \times 0.00630 \]
\[ = 0.00315 \text{ moles of } \text{Ba}_3(\text{PO}_4)_2\]
Thus, 0.00315 moles of the precipitate \(\text{Ba}_3(\text{PO}_4)_2\) are produced.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F163dc8ba-851f-4764-b6aa-51728fdb757c%2Fff57a681-9786-4281-82e8-c0661fee6b99%2Feujjboc_processed.png&w=3840&q=75)
Transcribed Image Text:**Balanced Chemical Reaction Problem**
**Question:**
Consider the reaction of 30.0 mL of 0.244 M BaI₂ with 20.0 mL of 0.315 M Na₃PO₄.
The balanced chemical equation for the reaction is:
\[ 3 \, \text{BaI}_2(aq) + 2 \, \text{Na}_3\text{PO}_4(aq) \rightarrow \text{Ba}_3(\text{PO}_4)_2(s) + 6 \, \text{NaI}(aq) \]
If you have 0.00630 moles of \(\text{Na}_3\text{PO}_4\), what quantity in moles of precipitate is produced if all the \(\text{Na}_3\text{PO}_4\) were consumed based on the balanced chemical equation?
**Answer:**
To solve this, first determine the moles of precipitate (\(\text{Ba}_3(\text{PO}_4)_2\)) that are formed according to the stoichiometric coefficients in the balanced equation. According to the equation, 2 moles of \(\text{Na}_3\text{PO}_4\) produce 1 mole of \(\text{Ba}_3(\text{PO}_4)_2\).
Given:
- Moles of \(\text{Na}_3\text{PO}_4\) = 0.00630 moles
Using stoichiometry:
\[ \text{Moles of } \text{Ba}_3(\text{PO}_4)_2 = \frac{1}{2} \times \text{Moles of Na}_3\text{PO}_4\]
\[ = \frac{1}{2} \times 0.00630 \]
\[ = 0.00315 \text{ moles of } \text{Ba}_3(\text{PO}_4)_2\]
Thus, 0.00315 moles of the precipitate \(\text{Ba}_3(\text{PO}_4)_2\) are produced.
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