A rigid container has both ethane gas (C2H6) and oxygen gas. partial pressure of ethane is 896.8 torr, and the partial pressure of oxygen gas is 2523.2 torr. The volume of the container is 5.84 L...
A rigid container has both ethane gas (C2H6) and oxygen gas.
partial pressure of ethane is 896.8 torr, and the partial pressure of oxygen gas is 2523.2 torr. The volume of the container is 5.84 L and the temperature of the gas mixture is 36.7 degrees Celsius. The mixture is then ignited and allowed to combust. After the combustion, the temperature is brought down to 22.9 degrees Celsius, and enough time is given to allow the water that is produced to condense to a liquid. What mass of water is produced? What is the total pressure in the container after the reaction and condensation of the water? Assume that the water is condensed and totally removed. The remaining gases occupy the entire 5.84 L container.
There are many steps to this solution. First, we use the information in the problem (and Dalton's Law of Partial Pressure) and the Ideal Gas Law to determine the number of moles of each gas present in the container.
We then compare the number of moles present to the moles required from the balanced chemical reaction for the combustion of ethane to determine which gas, if any is in excess and the amount of carbon dioxide produced..
We can then determine the pressure of the remaining gases by once again applying the Ideal Gas Law.
from the ideal gas law
Moles of ethane = PeV/RT = 0.271 mole
Moles of oxygen = PoV/RT = 0.762 mole
The balanced equation for combustion of ethane is
2C2H6 + 7O2 ---> 4CO2 + 6H20
We see from the number of moles of ethane and oxygen present that all of the oxygen will be consumed leaving behind 0.053 mole of unconsumed ethane and 0.435 mole of Carbon dioxide.
Thus the new pressure will be
Pethane = nRT/V = 167torr
Pco2 = nRT/V = 1376 torr.
Giving a total pressure of 1543 torr.