Three gases (8.00G of methane CH4, , 18.0G of ethane C2H6, , and an unknown amou

Question

Three gases (8.00G of methane CH4, , 18.0G of ethane C2H6, , and an unknown amount of propane, C3H8) were added to the same 10.0-L container. At 23.0 C, the total pressure in the container is 3.90ATM. Calculate the partial pressure of each gas in the container. Three gases (8.00G of methane CH4, , 18.0G of ethane C2H6, , and an unknown amount of propane, C3H8) were added to the same 10.0-L container. At 23.0 C, the total pressure in the container is 3.90ATM. Calculate the partial pressure of each gas in the container.

Explanation / Answer

Use the formula

PV = nRT
n total = 3.90 atm * 10.0 L / 0.0821 atm-L / mol-K * 296 K
= 1.604 moles

so, it becomes nCH4 + nC2H6 + nC3H8 = 1.604
nCH4 = 8 / 16 = 0.5mol
nC2H6 = 18 / 30 = 0.6mol

nC3H8 = 1.604 mol - ( 0.5+ 0.6 )mol= 0.504 mol
Weight of thepropane gas = 0.54 mol *44 g /mol = 23.76 g

PCH4 = 3.90 * 0.5 / 1.60 = 1.22 atm
PC2H6 = 3.90 * 0.6 / 1.604 = 1.46 atm
PC3H8 = 3.90 x 0.54 /1.604 = 1.313 atm answer

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