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 $\degree$C if the total pressure of the oxygen and propane is 1.00 atm? The molar mass of C₃H₆ is (42.08 g/mol.)

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Quizplus · Accepted Answer

To find the mass of cyclopropane gas, we first need to calculate the partial pressure of cyclopropane using Dalton's Law of Partial Pressures. Given a 1:4 mole ratio of cyclopropane to oxygen, the total moles of gas in the mixture is 5 moles (1 mole of cyclopropane + 4 moles of oxygen). Therefore, the partial pressure of cyclopropane is 1/5 of the total pressure, which is 1.00 atm, so the partial pressure of cyclopropane is 0.20 atm. Using the ideal gas law (PV = nRT), we can solve for the number of moles of cyclopropane (n). The volume (V) is 2.00 L, R is 0.0821 L·atm/(mol·K), and the temperature (T) must be converted to Kelvin: 23.5°C + 273.15 = 296.65 K. Plugging in the values: (0.20 atm)(2.00 L) = n(0.0821 L·atm/(mol·K))(296.65 K). Solving for n gives approximately 0.0165 moles of cyclopropane. Multiplying the moles by the molar mass of cyclopropane (42.08 g/mol) gives the mass: 0.0165 moles * 42.08 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