Solution Found!
Answer: A mixture of gases contains 0.31 mol CH4, 0.25 mol
Chapter 5, Problem 67P(choose chapter or problem)
A mixture of gases contains \(0.31\) mol \(\mathrm{CH}_{4}, 0.25\) mol \(\mathrm{C}_{2} \mathrm{H}_{6}\), and \(0.29\) mol \(\mathrm{C}_{3} \mathrm{H}_{8}\). The total pressure is \(1.50 \mathrm{~atm}\). Calculate the partial pressures of the gases.
Questions & Answers
QUESTION:
A mixture of gases contains \(0.31\) mol \(\mathrm{CH}_{4}, 0.25\) mol \(\mathrm{C}_{2} \mathrm{H}_{6}\), and \(0.29\) mol \(\mathrm{C}_{3} \mathrm{H}_{8}\). The total pressure is \(1.50 \mathrm{~atm}\). Calculate the partial pressures of the gases.
ANSWER:Step 1 of 3
The goal of the problem is to calculate the partial pressures of the gases.
Given:
Number of moles of \(\mathrm{CH}_{4}=0.31 \mathrm{~mol}\)
Number of moles of \(\mathrm{C}_{2} \mathrm{H}_{6}=0.25 \mathrm{~mol}\)
Number of moles of \(\mathrm{C}_{3} \mathrm{H}_{8}=0.29 \mathrm{~mol}\)
The total pressure of gases \(=1.50 \mathrm{~atm}\).