A mixture of cyclopropane gas (C₃H₆) and oxygen (O₂) in a 1.00 to 4.00 mole ratio is used as an anesthetic. What mass of cyclopropane gas is present in a 2.00 L flask at 23.5°C if the total pressure of the oxygen and propane is 1.00 atm? (The molar mass of C₃H₆ is 42.078 g/mol.)

Question

Quizplus · Accepted Answer

To find the mass of cyclopropane gas in the mixture, we first need to determine the partial pressure of cyclopropane using Dalton's Law of Partial Pressures. Given a 1:4 mole ratio of cyclopropane to oxygen, for every 5 moles of gas, 1 mole is cyclopropane and 4 moles are oxygen. This means cyclopropane constitutes 1/5 of the total moles, and thus its partial pressure is 1/5 of the total pressure (1.00 atm), which is 0.20 atm.Next, we use the ideal gas law, PV = nRT, to find the number of moles of cyclopropane. Rearranging the formula to solve for n (moles), we get n = PV/RT. The values are as follows: P = 0.20 atm, V = 2.00 L, R = 0.0821 L·atm/(mol·K), and T = 23.5°C + 273.15 = 296.65 K.Substituting these values in gives n = (0.20 atm * 2.00 L) / (0.0821 L·atm/(mol·K) * 296.65 K) = 0.0164 moles of cyclopropane.Finally, to find the mass of cyclopropane, we multiply the number of moles by the molar mass: 0.0164 moles * 42.078 g/mol = 0.69 g.

A Mixture of Cyclopropane Gas (C3H6) and Oxygen (O2) in a 1.00

A Mixture of Cyclopropane Gas (C₃H₆) and Oxygen (O₂) in a 1.00