Chapter 5: Problem 119
Equal weights of methane and hydrogen are mixed in an empty container at \(25^{\circ} \mathrm{C}\). The fraction of the total pressure exerted by hydrogen is (1) \(\frac{1}{2}\) (2) \(\frac{8}{9}\) (3) \(\frac{16}{17}\) (4) \(\frac{1}{9}\)
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
molar mass calculation
12 (C) + 4*1 (H) = 16 g/mol.
Similarly, for hydrogen gas (H₂), composed of two hydrogen atoms, the molar mass is: 2*1 (H) = 2 g/mol.
partial pressure
P_total = P1 + P2 + ... + Pn
where P_total is the total pressure, and P1, P2,..., Pn are the partial pressures of gases 1, 2,..., n, respectively. In the exercise, we determined the partial pressure fraction exerted by hydrogen. Since the mole fraction of hydrogen is the ratio of the moles of hydrogen to the total moles of gas, we used this ratio to find the fraction of the total pressure.
ideal gas law
P = Pressure of the gas
V = Volume of the gas
n = Number of moles of the gas
R = Universal gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin (K)
In the given exercise, although we did not directly use the Ideal Gas Law equation, concepts from it were essential. By calculating the total number of moles, we implicitly relied on the relationships described by this law.
stoichiometry
- Convert given masses to moles using molar masses.
- Use the mole ratio to understand the relationship between different components.
- Apply the results to find properties like partial pressures or total volumes.