A mixture of methane gas, CH_4(g). and propane gas, C_3H_8(Sf)> has a pressure o

Question

A mixture of methane gas, CH_4(g). and propane gas, C_3H_8(Sf)> has a pressure of 0.5642 atm when placed in a sealed container. The complete combustion of the mixture to carbon dioxide gas, CO_2(g), and water vapor, H_2O(g), was achieved by adding exactly enough oxygen gas, O_2(g), to the container. The pressure of the product mixture in the sealed container is 2.302 atm. Calculate the mole fraction of methane in the initial mixture assuming the temperature and volume remain constant. Xch4 Number

Explanation / Answer

Answer: According to question , Here we have to use the formula Partial pressure = X Ptotal

Now for Methane we have partial pressure = 0.5642 / 2 [ it is the pressure of mixture hence for CH4 it is half ]

= 0.2821 atm

and P total = 2.302 atm

Hence we get , 0.2821 = XCH4 * 2.302

XCH4 = 0.1225

Hecne the required answer is 0.1225 .

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