How many grams of Ag₂CO3 will precipitate when excess K₂CO3 solution is added to 57.0 mL of 0.658 M AgNO3 solution? 2AgNO3(aq) + K2CO3(aq) → Ag2 CO3(s) + 2KNO3(aq) g
How many grams of Ag₂CO3 will precipitate when excess K₂CO3 solution is added to 57.0 mL of 0.658 M AgNO3 solution? 2AgNO3(aq) + K2CO3(aq) → Ag2 CO3(s) + 2KNO3(aq) g
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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![### Chemistry Problem: Precipitation of Silver Carbonate
**Problem Statement:**
How many grams of \( \text{Ag}_2\text{CO}_3 \) will precipitate when excess \( \text{K}_2\text{CO}_3 \) solution is added to 57.0 mL of 0.658 M \( \text{AgNO}_3 \) solution?
**Chemical Equation:**
\[ 2 \text{AgNO}_3 (\text{aq}) + \text{K}_2\text{CO}_3 (\text{aq}) \rightarrow \text{Ag}_2\text{CO}_3 (\text{s}) + 2 \text{KNO}_3 (\text{aq}) \]
**Answer Placeholder:**
\[ \boxed{} \text{ g} \]
**Explanation:**
1. **Identify the Limiting Reactant:**
- Given that \( \text{K}_2\text{CO}_3 \) is in excess, \( \text{AgNO}_3 \) will be the limiting reactant.
2. **Calculate the Moles of \( \text{AgNO}_3 \):**
- Volume of \( \text{AgNO}_3 \) solution = 57.0 mL = 0.0570 L
- Molarity of \( \text{AgNO}_3 \) (M) = 0.658 M
- Moles of \( \text{AgNO}_3 \) = Molarity × Volume
\[ \text{Moles of } \text{AgNO}_3 = 0.658 \, \text{M} \times 0.0570 \, \text{L} \]
3. **Calculate the Moles of \( \text{Ag}_2\text{CO}_3 \) Formed:**
- From the balanced chemical equation, 2 moles of \( \text{AgNO}_3 \) form 1 mole of \( \text{Ag}_2\text{CO}_3 \).
- Therefore, moles of \( \text{Ag}_2\text{CO}_3 = \frac{1}{2} \times \text{moles of } \text{AgNO}_3 \)
\[ \text{Moles of } \text{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a354f1c-cdfd-41e5-a863-9a86a504af1c%2F20e1a2bc-d2fd-4ca4-ae8f-715a7c74e108%2Fz32ghus_processed.png&w=3840&q=75)
Transcribed Image Text:### Chemistry Problem: Precipitation of Silver Carbonate
**Problem Statement:**
How many grams of \( \text{Ag}_2\text{CO}_3 \) will precipitate when excess \( \text{K}_2\text{CO}_3 \) solution is added to 57.0 mL of 0.658 M \( \text{AgNO}_3 \) solution?
**Chemical Equation:**
\[ 2 \text{AgNO}_3 (\text{aq}) + \text{K}_2\text{CO}_3 (\text{aq}) \rightarrow \text{Ag}_2\text{CO}_3 (\text{s}) + 2 \text{KNO}_3 (\text{aq}) \]
**Answer Placeholder:**
\[ \boxed{} \text{ g} \]
**Explanation:**
1. **Identify the Limiting Reactant:**
- Given that \( \text{K}_2\text{CO}_3 \) is in excess, \( \text{AgNO}_3 \) will be the limiting reactant.
2. **Calculate the Moles of \( \text{AgNO}_3 \):**
- Volume of \( \text{AgNO}_3 \) solution = 57.0 mL = 0.0570 L
- Molarity of \( \text{AgNO}_3 \) (M) = 0.658 M
- Moles of \( \text{AgNO}_3 \) = Molarity × Volume
\[ \text{Moles of } \text{AgNO}_3 = 0.658 \, \text{M} \times 0.0570 \, \text{L} \]
3. **Calculate the Moles of \( \text{Ag}_2\text{CO}_3 \) Formed:**
- From the balanced chemical equation, 2 moles of \( \text{AgNO}_3 \) form 1 mole of \( \text{Ag}_2\text{CO}_3 \).
- Therefore, moles of \( \text{Ag}_2\text{CO}_3 = \frac{1}{2} \times \text{moles of } \text{AgNO}_3 \)
\[ \text{Moles of } \text{
![**Problem Statement:**
*How many mL of 0.535 M HI are needed to dissolve 7.98 g of BaCO₃?*
**Chemical Reaction:**
\[ 2\text{HI}(aq) + \text{BaCO}_3(s) \rightarrow \text{BaI}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g) \]
**Input Field:**
\[ \underline{\hspace{1cm}} \text{ mL} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7a354f1c-cdfd-41e5-a863-9a86a504af1c%2F20e1a2bc-d2fd-4ca4-ae8f-715a7c74e108%2Fird01c_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
*How many mL of 0.535 M HI are needed to dissolve 7.98 g of BaCO₃?*
**Chemical Reaction:**
\[ 2\text{HI}(aq) + \text{BaCO}_3(s) \rightarrow \text{BaI}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g) \]
**Input Field:**
\[ \underline{\hspace{1cm}} \text{ mL} \]
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