How many moles of precipitate will be formed when 69.9 mL of 0.300 M AGNO: is reacted with excess Cal2 in the following chemical reaction? 2 AgNO: (aq) + Cal2 (aq) → 2 Agl (s) + Ca(NO:)2 (aq)
How many moles of precipitate will be formed when 69.9 mL of 0.300 M AGNO: is reacted with excess Cal2 in the following chemical reaction? 2 AgNO: (aq) + Cal2 (aq) → 2 Agl (s) + Ca(NO:)2 (aq)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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![**Question 9 of 16**
**Problem Statement:**
How many moles of precipitate will be formed when 69.9 mL of 0.300 M AgNO₃ is reacted with excess CaI₂ in the following chemical reaction?
\[ 2 \, \text{AgNO}_3 \, (\text{aq}) + \text{CaI}_2 \, (\text{aq}) \rightarrow 2 \, \text{AgI} \, (\text{s}) + \text{Ca(NO}_3)_2 \, (\text{aq}) \]
**Explanation:**
This problem involves a chemical reaction where silver nitrate (AgNO₃) reacts with calcium iodide (CaI₂) to form silver iodide (AgI) as a precipitate and calcium nitrate (Ca(NO₃)₂).
- The solution has a volume of 69.9 mL and a concentration of 0.300 M for AgNO₃.
- The reaction between AgNO₃ and CaI₂ follows a 2:1 stoichiometry, meaning two moles of AgNO₃ produce two moles of AgI precipitate.
To find the moles of precipitate (AgI), use:
1. **Convert the volume from mL to L**: \( 69.9 \, \text{mL} = 0.0699 \, \text{L} \)
2. **Calculate moles of AgNO₃ using Molarity**:
\[
\text{Moles of AgNO}_3 = 0.0699 \, \text{L} \times 0.300 \, \text{mol/L} = 0.02097 \, \text{mol}
\]
3. **Stoichiometry**: Since 2 moles of AgNO₃ produce 2 moles of AgI, the moles of AgI will also be 0.02097.
Thus, 0.02097 moles of AgI precipitate will be formed.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0650aa8f-2b62-480a-8217-30989ebe5695%2F7ccc9516-3c63-44b3-a6f9-1ecda294800a%2Fpfhqxpd.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 9 of 16**
**Problem Statement:**
How many moles of precipitate will be formed when 69.9 mL of 0.300 M AgNO₃ is reacted with excess CaI₂ in the following chemical reaction?
\[ 2 \, \text{AgNO}_3 \, (\text{aq}) + \text{CaI}_2 \, (\text{aq}) \rightarrow 2 \, \text{AgI} \, (\text{s}) + \text{Ca(NO}_3)_2 \, (\text{aq}) \]
**Explanation:**
This problem involves a chemical reaction where silver nitrate (AgNO₃) reacts with calcium iodide (CaI₂) to form silver iodide (AgI) as a precipitate and calcium nitrate (Ca(NO₃)₂).
- The solution has a volume of 69.9 mL and a concentration of 0.300 M for AgNO₃.
- The reaction between AgNO₃ and CaI₂ follows a 2:1 stoichiometry, meaning two moles of AgNO₃ produce two moles of AgI precipitate.
To find the moles of precipitate (AgI), use:
1. **Convert the volume from mL to L**: \( 69.9 \, \text{mL} = 0.0699 \, \text{L} \)
2. **Calculate moles of AgNO₃ using Molarity**:
\[
\text{Moles of AgNO}_3 = 0.0699 \, \text{L} \times 0.300 \, \text{mol/L} = 0.02097 \, \text{mol}
\]
3. **Stoichiometry**: Since 2 moles of AgNO₃ produce 2 moles of AgI, the moles of AgI will also be 0.02097.
Thus, 0.02097 moles of AgI precipitate will be formed.
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