Posted by Fred on .
A mixture of ethyne ( C2H2 ) and methane ( CH4 ) occupied a certain volume at a total pressure of 16.8 kPa. Upon burning the sample to form CO2 and H2O, the CO2 was collected and its pressure found to be 25.6 kPa in the same volume and at the same temperature as the original mixture. What percent of the original mixture was methane?
The reactions are:
2C2H2(g) + 5O2(g) —> 4CO2(g) + 2H2O(L)
CH4 + 2O2 —> CO2(g) + 2H2O(L)
Let moles of C2H2 = 2x, moles CH4 = y. Then,
2xC2H2(g) + 5xO2(g) —> 4xCO2(g) + 2xH2O(L)
yCH4 + 2yO2 —> yCO2(g) + 2yH2O(L)
Original moles of gas (C2H2 + CH4) = 2x + y
Final moles of gas (CO2) = 4x + y
(4x + y) / (2x + y) = 25.6 / 16.8 = ____?
Solve for the ratio x/y
Assign the value y = 1 and solve for x.
Determine the ratio:
y / (x + y)
The above ratio times 100 is the % of methane.