A 30.0 L liquefied petroleum gas cylinder consists of 55% propane by mass and 45% of butane. If the total mass of the liquefied gases is 23.0 kg, then what is the partial pressure exerted by propane in the cylinder at 25.0˚C? Molar mass of propane is 44.096 g/mol.

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To find the partial pressure exerted by propane, we first determine the mass of propane in the mixture, then convert that mass to moles, and finally use the ideal gas law to find the pressure.1. Calculate the mass of propane:$$ \text{Mass of propane} = 55\% \times 23.0\, \text{kg} = 0.55 \times 23.0 = 12.65\, \text{kg} $$$$ = 12650\, \text{g} $$ (since 1 kg = 1000 g)2. Convert the mass of propane to moles:$$ \text{Moles of propane} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{12650\, \text{g}}{44.096\, \text{g/mol}} \approx 286.9\, \text{mol} $$3. Use the ideal gas law to find the partial pressure of propane. The ideal gas law is $ 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/(mol·K)), and $ T $ is the temperature in Kelvin.$$ T = 25.0˚C = 25.0 + 273.15 = 298.15\, K $$$$ P = \frac{nRT}{V} $$$$ P = \frac{286.9\, \text{mol} \times 0.0821\, \text{L·atm/(mol·K)} \times 298.15\, K}{30.0\, L} $$$$ P \approx 24.1\, \text{atm} $$Therefore, the partial pressure exerted by propane in the cylinder at 25.0˚C is approximately 24.1 atm.

A 300 L Liquefied Petroleum Gas Cylinder Consists of 55% Propane