How many grams of AgCl will be formed when 60.0 mL of 0.500 M AgNO, is completely reacted according to the balanced chemical reaction: FeCl₂(aq) + 3 AgNO,(aq) 3 AgCl(s) + Fe(NO₂),(aq)
How many grams of AgCl will be formed when 60.0 mL of 0.500 M AgNO, is completely reacted according to the balanced chemical reaction: FeCl₂(aq) + 3 AgNO,(aq) 3 AgCl(s) + Fe(NO₂),(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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can you help me plug it in the right form i keep doing it and its wrong
![This educational example addresses the calculation of how many grams of AgCl can be formed when a solution of AgNO₃ is reacted according to a balanced chemical equation:
**Chemical Reaction:**
\[ \text{FeCl}_3 (aq) + 3 \text{AgNO}_3 (aq) \rightarrow 3 \text{AgCl} (s) + \text{Fe(NO}_3)_3 (aq) \]
**Given:**
- Volume of AgNO₃: 60.0 mL
- Molarity of AgNO₃: 0.500 M
**Solution Steps:**
1. **Convert mL AgNO₃ to mol AgCl:**
\[
\text{Starting Amount: } 60.0 \text{ mL AgNO}_3 \times \left( \frac{0.500 \text{ mol AgCl}}{1 \text{ mol AgNO}_3} \right) \times \left( \frac{143.32 \text{ g AgCl}}{1 \text{ mol AgCl}} \right) \times \left( \frac{1 \text{ mol AgCl}}{3 \text{ mol AgCl}} \right)
\]
2. **Calculate the Grams of AgCl Formed:**
\[
= 4.30 \text{ g AgCl}
\]
**Calculation Path:**
- \(60.0 \text{ mL AgNO}_3\) is multiplied by the molarity (0.500 M) to convert to moles of AgNO₃.
- This conversion factor leads to moles of AgCl via the stoichiometric coefficients.
- Moles of AgCl are then converted to grams using the molar mass (143.32 g/mol).
**Diagram:**
- The equation setup is shown using conversion factors and stoichiometry, with intermediate values corresponding to each step, leading to the final answer of 4.30 grams of AgCl.
**Interactive Elements:**
- Buttons and fields such as "ADD FACTOR", "DELETE", "ANSWER", and "RESET" aid in visualizing and calculating.
- Various numerical values are shown for active selection during calculations.
This example illustrates the application of stoichiometry in chemical reactions, highlighting the systematic conversion process from volume and concentration to the mass of a product.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F574bf9f3-609e-4870-b103-4ad8e97b08b1%2Fde570d3a-26a2-4be1-951b-c5496f77152d%2Fhlt33mn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:This educational example addresses the calculation of how many grams of AgCl can be formed when a solution of AgNO₃ is reacted according to a balanced chemical equation:
**Chemical Reaction:**
\[ \text{FeCl}_3 (aq) + 3 \text{AgNO}_3 (aq) \rightarrow 3 \text{AgCl} (s) + \text{Fe(NO}_3)_3 (aq) \]
**Given:**
- Volume of AgNO₃: 60.0 mL
- Molarity of AgNO₃: 0.500 M
**Solution Steps:**
1. **Convert mL AgNO₃ to mol AgCl:**
\[
\text{Starting Amount: } 60.0 \text{ mL AgNO}_3 \times \left( \frac{0.500 \text{ mol AgCl}}{1 \text{ mol AgNO}_3} \right) \times \left( \frac{143.32 \text{ g AgCl}}{1 \text{ mol AgCl}} \right) \times \left( \frac{1 \text{ mol AgCl}}{3 \text{ mol AgCl}} \right)
\]
2. **Calculate the Grams of AgCl Formed:**
\[
= 4.30 \text{ g AgCl}
\]
**Calculation Path:**
- \(60.0 \text{ mL AgNO}_3\) is multiplied by the molarity (0.500 M) to convert to moles of AgNO₃.
- This conversion factor leads to moles of AgCl via the stoichiometric coefficients.
- Moles of AgCl are then converted to grams using the molar mass (143.32 g/mol).
**Diagram:**
- The equation setup is shown using conversion factors and stoichiometry, with intermediate values corresponding to each step, leading to the final answer of 4.30 grams of AgCl.
**Interactive Elements:**
- Buttons and fields such as "ADD FACTOR", "DELETE", "ANSWER", and "RESET" aid in visualizing and calculating.
- Various numerical values are shown for active selection during calculations.
This example illustrates the application of stoichiometry in chemical reactions, highlighting the systematic conversion process from volume and concentration to the mass of a product.
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