When copper metal is added to nitric acid, the following reaction takes place Cu (s) + 4 HNO3 (aq) → Cu(NO3)2 (aq) + 2 H2O (I) + 2 NO2 (g) Calculate the volume in liters of NO2 gas collected over water at 25.0 °C when 3.29 g of copper is added to excess nitric acid if the total pressure is 726.0 mmHg. The vapor pressure of water at 25.0 °C is 23.8 mm Hg.
When copper metal is added to nitric acid, the following reaction takes place Cu (s) + 4 HNO3 (aq) → Cu(NO3)2 (aq) + 2 H2O (I) + 2 NO2 (g) Calculate the volume in liters of NO2 gas collected over water at 25.0 °C when 3.29 g of copper is added to excess nitric acid if the total pressure is 726.0 mmHg. The vapor pressure of water at 25.0 °C is 23.8 mm Hg.
Chemistry
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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
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![**Title:** Copper and Nitric Acid Reaction and NO₂ Gas Collection
**Introduction:**
When copper metal reacts with nitric acid, a chemical reaction occurs that produces copper(II) nitrate, water, and nitrogen dioxide gas. This lesson will guide you through the process of calculating the volume of NO₂ gas generated in this reaction under given conditions.
**Reaction Description:**
The chemical equation for the reaction between copper and nitric acid is:
\[ \text{Cu (s) + 4 HNO}_3\, (\text{aq}) \rightarrow \text{Cu(NO}_3\text{)}_2\, (\text{aq}) + 2 \text{H}_2\text{O (l)} + 2 \text{NO}_2\, (\text{g}) \]
**Problem Statement:**
Calculate the volume in liters of NO₂ gas collected over water at 25.0 °C when 3.29 g of copper is added to excess nitric acid if the total pressure is 726.0 mmHg. The vapor pressure of water at 25.0 °C is 23.8 mmHg.
**Steps to Solve the Problem:**
1. **Determine the moles of copper:**
- The molar mass of copper (Cu) is 63.55 g/mol.
- Moles of Cu = \(\frac{\text{Mass of Cu}}{\text{Molar mass of Cu}}\) = \(\frac{3.29\, \text{g}}{63.55\, \text{g/mol}}\) = 0.0518 mol of Cu.
2. **Use the stoichiometry of the reaction:**
- According to the balanced equation, 1 mole of Cu produces 2 moles of NO₂.
- Moles of NO₂ = 0.0518 mol Cu × \(\frac{2\, \text{mol NO}_2}{1\, \text{mol Cu}}\) = 0.1036 mol NO₂.
3. **Correct for the vapor pressure of water:**
- Partial pressure of NO₂, \(P_{\text{NO}_2}\):
\(P_{\text{NO}_2} = P_{\text{total}} - P_{\text{H}_2\text{O}}\)
\(P_{\text{NO}_2} = 726.0\, \text{mmHg](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F44982666-d339-4885-b9e9-36706f6e8dd1%2Fd522405d-5ac9-47c4-81fb-71dcd441e930%2Fcl74zn_processed.png&w=3840&q=75)
Transcribed Image Text:**Title:** Copper and Nitric Acid Reaction and NO₂ Gas Collection
**Introduction:**
When copper metal reacts with nitric acid, a chemical reaction occurs that produces copper(II) nitrate, water, and nitrogen dioxide gas. This lesson will guide you through the process of calculating the volume of NO₂ gas generated in this reaction under given conditions.
**Reaction Description:**
The chemical equation for the reaction between copper and nitric acid is:
\[ \text{Cu (s) + 4 HNO}_3\, (\text{aq}) \rightarrow \text{Cu(NO}_3\text{)}_2\, (\text{aq}) + 2 \text{H}_2\text{O (l)} + 2 \text{NO}_2\, (\text{g}) \]
**Problem Statement:**
Calculate the volume in liters of NO₂ gas collected over water at 25.0 °C when 3.29 g of copper is added to excess nitric acid if the total pressure is 726.0 mmHg. The vapor pressure of water at 25.0 °C is 23.8 mmHg.
**Steps to Solve the Problem:**
1. **Determine the moles of copper:**
- The molar mass of copper (Cu) is 63.55 g/mol.
- Moles of Cu = \(\frac{\text{Mass of Cu}}{\text{Molar mass of Cu}}\) = \(\frac{3.29\, \text{g}}{63.55\, \text{g/mol}}\) = 0.0518 mol of Cu.
2. **Use the stoichiometry of the reaction:**
- According to the balanced equation, 1 mole of Cu produces 2 moles of NO₂.
- Moles of NO₂ = 0.0518 mol Cu × \(\frac{2\, \text{mol NO}_2}{1\, \text{mol Cu}}\) = 0.1036 mol NO₂.
3. **Correct for the vapor pressure of water:**
- Partial pressure of NO₂, \(P_{\text{NO}_2}\):
\(P_{\text{NO}_2} = P_{\text{total}} - P_{\text{H}_2\text{O}}\)
\(P_{\text{NO}_2} = 726.0\, \text{mmHg
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