How many grams of AgCI will be formed when 60 mL of 0.500M AGNO, is completely reacted according to the balanced chemical reaction: FeCl,(aq) + 3 AGNO,(aq) - 3 AGCI(s) + Fe(NO,)2(aq) STARTING AMOUNT
How many grams of AgCI will be formed when 60 mL of 0.500M AGNO, is completely reacted according to the balanced chemical reaction: FeCl,(aq) + 3 AGNO,(aq) - 3 AGCI(s) + Fe(NO,)2(aq) STARTING AMOUNT
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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Please use dimensional analysis and only use the numbers that are available in this picture. note that 0.03 is not an option.
![**Stoichiometry Calculation Practice**
**Question 4 of 10**
**Problem Statement:**
Determine how many grams of AgCl will be formed when 60 mL of 0.500 M AgNO₃ is completely reacted according to the balanced chemical reaction:
\[ \text{FeCl}_3(\text{aq}) + 3 \text{AgNO}_3(\text{aq}) \rightarrow 3 \text{AgCl}(\text{s}) + \text{Fe(NO}_3\text{)}_3(\text{aq}) \]
**Calculation Setup:**
- Calculate the amount of AgCl using the provided stoichiometry.
**Calculation Box:**
1. **Starting Amount:**
- Input fields for initial quantities.
2. **Multiplication and Conversion:**
- Factor inputs for conversion (e.g., moles to grams).
**Available Factors for Calculation:**
- \(6.022 \times 10^{23}\) (Avogadro's number)
- Conversion constants, molar masses, and volumes:
| Value | Description |
|---------|-------------------|
| 12.9 | |
| 1000 | |
| 169.88 | Molar mass of AgCl (g/mol) |
| 143.32 | |
| 1.70 | |
| 1.43 | |
| 1 | |
| 60 | Volume (mL) |
| 0.500 | Molarity (M) |
| 3 | Stoichiometric coefficient |
| 0.001 | |
| 4.30 | |
- Units for Conversion:
- g AgNO₃, mol AgCl, g AgCl, mL AgCl, M AgNO₃, mL AgNO₃, mol AgNO₃, L AgCl, L AgNO₃
**Interactive Features:**
- **Add Factor**: Select multiplication factors needed for conversion.
- **Answer Input**: Box for entering the final calculated amount.
- **Reset**: Option to clear the current calculation and restart.
**Instructions:**
1. Use the balanced equation to find the stoichiometric ratio.
2. Convert mL of AgNO₃ to](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F241f8404-7873-49cc-ad9f-2100b3d757fc%2F0e0daa49-2b82-41ef-8397-a4111526bdd8%2Fp9gu2tn_processed.png&w=3840&q=75)
Transcribed Image Text:**Stoichiometry Calculation Practice**
**Question 4 of 10**
**Problem Statement:**
Determine how many grams of AgCl will be formed when 60 mL of 0.500 M AgNO₃ is completely reacted according to the balanced chemical reaction:
\[ \text{FeCl}_3(\text{aq}) + 3 \text{AgNO}_3(\text{aq}) \rightarrow 3 \text{AgCl}(\text{s}) + \text{Fe(NO}_3\text{)}_3(\text{aq}) \]
**Calculation Setup:**
- Calculate the amount of AgCl using the provided stoichiometry.
**Calculation Box:**
1. **Starting Amount:**
- Input fields for initial quantities.
2. **Multiplication and Conversion:**
- Factor inputs for conversion (e.g., moles to grams).
**Available Factors for Calculation:**
- \(6.022 \times 10^{23}\) (Avogadro's number)
- Conversion constants, molar masses, and volumes:
| Value | Description |
|---------|-------------------|
| 12.9 | |
| 1000 | |
| 169.88 | Molar mass of AgCl (g/mol) |
| 143.32 | |
| 1.70 | |
| 1.43 | |
| 1 | |
| 60 | Volume (mL) |
| 0.500 | Molarity (M) |
| 3 | Stoichiometric coefficient |
| 0.001 | |
| 4.30 | |
- Units for Conversion:
- g AgNO₃, mol AgCl, g AgCl, mL AgCl, M AgNO₃, mL AgNO₃, mol AgNO₃, L AgCl, L AgNO₃
**Interactive Features:**
- **Add Factor**: Select multiplication factors needed for conversion.
- **Answer Input**: Box for entering the final calculated amount.
- **Reset**: Option to clear the current calculation and restart.
**Instructions:**
1. Use the balanced equation to find the stoichiometric ratio.
2. Convert mL of AgNO₃ to
Expert Solution
Step 1
The balanced reaction taking place is given as,
Given : Concentration of AgNO3 = 0.500 M
And volume of AgNO3 solution = 60 mL = 0.060 L (since 1 L = 1000 mL)
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