Similar to For practice 6.12)In the reaction shown here, 955 mL of O₂ forms at P = 0.950 atm and T-308 K. H s of Ag2O (molar mass = 231.8 g/mol) decomposed? O(s) --> 4 Ag(s) + O2(g) 15 g 574 g

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Chapter1: Chemical Foundations
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### Problem Description:
**(6.7: Similar to For practice 6.12)** In the reaction shown here, 955 mL of O₂ forms at P = 0.950 atm and T = 308 K. How many grams of Ag₂O (molar mass = 231.8 g/mol) decomposed?

\[ 2Ag_2O(s) \rightarrow 4Ag(s) + O_2(g) \]

### Possible Answers:
- **☐ 1.15 g**
- **☐ 0.574 g**
- **☐ 8.32 g**
- **☐ 16.6 g**

### Solution Steps:
1. **Calculate the moles of O₂ formed** using the Ideal Gas Law equation PV = nRT, where:
   - P = 0.950 atm
   - V = 955 mL = 0.955 L (Note: 1 L = 1000 mL)
   - T = 308 K
   - R = 0.0821 L·atm/K·mol (Ideal Gas Constant)
   
   Rearrange to find \( n \) (number of moles of O₂):
   \[ n = \frac{PV}{RT} \]

2. **Use stoichiometry to relate moles of O₂ to moles of Ag₂O**. According to the balanced chemical equation:
   \[ 2Ag_2O(s) \rightarrow 4Ag(s) + O_2(g) \]
   - 1 mole of O₂ is produced from 2 moles of Ag₂O.

3. **Calculate the mass of Ag₂O needed**.
   - Use the molar mass of Ag₂O (231.8 g/mol) and the number of moles of Ag₂O found in the stoichiometry step.

### Detailed Explanations:
- **Graphs/Diagrams**: There are no graphs or diagrams in this problem.

### Educational Objective:
The goal is to teach students how to use the Ideal Gas Law in combination with stoichiometry to solve for the mass of a reactant in a chemical reaction.
Transcribed Image Text:### Problem Description: **(6.7: Similar to For practice 6.12)** In the reaction shown here, 955 mL of O₂ forms at P = 0.950 atm and T = 308 K. How many grams of Ag₂O (molar mass = 231.8 g/mol) decomposed? \[ 2Ag_2O(s) \rightarrow 4Ag(s) + O_2(g) \] ### Possible Answers: - **☐ 1.15 g** - **☐ 0.574 g** - **☐ 8.32 g** - **☐ 16.6 g** ### Solution Steps: 1. **Calculate the moles of O₂ formed** using the Ideal Gas Law equation PV = nRT, where: - P = 0.950 atm - V = 955 mL = 0.955 L (Note: 1 L = 1000 mL) - T = 308 K - R = 0.0821 L·atm/K·mol (Ideal Gas Constant) Rearrange to find \( n \) (number of moles of O₂): \[ n = \frac{PV}{RT} \] 2. **Use stoichiometry to relate moles of O₂ to moles of Ag₂O**. According to the balanced chemical equation: \[ 2Ag_2O(s) \rightarrow 4Ag(s) + O_2(g) \] - 1 mole of O₂ is produced from 2 moles of Ag₂O. 3. **Calculate the mass of Ag₂O needed**. - Use the molar mass of Ag₂O (231.8 g/mol) and the number of moles of Ag₂O found in the stoichiometry step. ### Detailed Explanations: - **Graphs/Diagrams**: There are no graphs or diagrams in this problem. ### Educational Objective: The goal is to teach students how to use the Ideal Gas Law in combination with stoichiometry to solve for the mass of a reactant in a chemical reaction.
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