A 20.0 L container at 303 K holds a mixture of two gases with a total pressure of 5.00 atm. If there are 1.00 mol of Gas A in the mixture, What quantity in moles of Gas B are present?
A 20.0 L container at 303 K holds a mixture of two gases with a total pressure of 5.00 atm. If there are 1.00 mol of Gas A in the mixture, What quantity in moles of Gas B are present?
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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![**Question 14 of 21**
**Educational Content**
Title: Calculating the Moles of Gases in a Mixture
**Problem Statement:**
A 20.0 L container at 303 K holds a mixture of two gases with a total pressure of 5.00 atm. If there are 1.00 mol of Gas A in the mixture, what quantity in moles of Gas B are present?
**Explanation:**
In this problem, we are given the following details:
- Volume of the container (V) = 20.0 L
- Temperature (T) = 303 K
- Total pressure (P) = 5.00 atm
- Moles of Gas A (n_A) = 1.00 mol
To find the moles of Gas B (n_B), we should use the ideal gas law, and the concept of partial pressures can help.
From the ideal gas law:
\[ PV = nRT \]
Where:
- \( P \) is the pressure
- \( V \) is the volume
- \( n \) is the number of moles
- \( R \) is the ideal gas constant (0.0821 L·atm·K⁻¹·mol⁻¹)
- \( T \) is the temperature in Kelvin
First, calculate the total moles of the gas mixture \( (n_{total}) \):
\[ P \cdot V = n_{total} \cdot R \cdot T \]
\[ 5.00 \, atm \cdot 20.0 \, L = n_{total} \cdot 0.0821 \, L · atm · K^{-1} · mol^{-1} \cdot 303 \, K \]
\[ 100 = n_{total} \cdot 24.8763 \]
\[ n_{total} = \frac{100}{24.8763} \]
\[ n_{total} \approx 4.02 \, mol \]
Now, knowing that \( n_{total} = n_A + n_B \):
\[ 4.02 \, mol = 1.00 \, mol + n_B \]
\[ n_B = 4.02 - 1.00 \]
\[ n_B = 3.02 \, mol \]
Therefore, the quantity of Gas B present in the mixture is approximately 3.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6895f9cd-8521-4c9a-bc62-df11c4ac3cae%2F8991a9ec-82a9-47e7-aa96-eb673e89b717%2Ffo8pwc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 14 of 21**
**Educational Content**
Title: Calculating the Moles of Gases in a Mixture
**Problem Statement:**
A 20.0 L container at 303 K holds a mixture of two gases with a total pressure of 5.00 atm. If there are 1.00 mol of Gas A in the mixture, what quantity in moles of Gas B are present?
**Explanation:**
In this problem, we are given the following details:
- Volume of the container (V) = 20.0 L
- Temperature (T) = 303 K
- Total pressure (P) = 5.00 atm
- Moles of Gas A (n_A) = 1.00 mol
To find the moles of Gas B (n_B), we should use the ideal gas law, and the concept of partial pressures can help.
From the ideal gas law:
\[ PV = nRT \]
Where:
- \( P \) is the pressure
- \( V \) is the volume
- \( n \) is the number of moles
- \( R \) is the ideal gas constant (0.0821 L·atm·K⁻¹·mol⁻¹)
- \( T \) is the temperature in Kelvin
First, calculate the total moles of the gas mixture \( (n_{total}) \):
\[ P \cdot V = n_{total} \cdot R \cdot T \]
\[ 5.00 \, atm \cdot 20.0 \, L = n_{total} \cdot 0.0821 \, L · atm · K^{-1} · mol^{-1} \cdot 303 \, K \]
\[ 100 = n_{total} \cdot 24.8763 \]
\[ n_{total} = \frac{100}{24.8763} \]
\[ n_{total} \approx 4.02 \, mol \]
Now, knowing that \( n_{total} = n_A + n_B \):
\[ 4.02 \, mol = 1.00 \, mol + n_B \]
\[ n_B = 4.02 - 1.00 \]
\[ n_B = 3.02 \, mol \]
Therefore, the quantity of Gas B present in the mixture is approximately 3.
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