To solve this problem, we will use the ideal gas law and the stoichiometry of the chemical reaction.
Step 1: Write the balanced chemical equation for the combustion of methane (CH4) and ethane (C2H6) in air, which produces carbon dioxide (CO2) and water (H2O):
CH4 + 2O2 -> CO2 + 2H2O
C2H6 + 7O2 -> 2CO2 + 6H2O
Step 2: Calculate the moles of CO2 produced using the ideal gas law. Firstly, we need to convert the pressures from mmHg to atm:
294 mmHg = 0.387 atm (by dividing by 760 mmHg/atm)
351 mmHg = 0.461 atm
Step 3: Calculate the ratio of moles of CO2 produced to moles of the original gas mixture. To do this, we need to find the moles of CO2 and the moles of the original gas mixture (CH4 and C2H6).
For CO2:
PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature (which is constant in this case).
Since the pressure and temperature are the same for both CO2 and the original mixture, we can write:
0.387 atm * V = nCO2 * R * T
For the original gas mixture:
Ptotal * V = ntotal * R * T, where Ptotal is the total pressure of the mixture, V is the volume, ntotal is the number of moles of the mixture, R is the ideal gas constant, and T is the temperature (which is constant in this case).
Since the pressure and temperature are the same for both CO2 and the original mixture, we can write:
0.387 atm * V = ntotal * R * T
Step 4: Calculate the moles of CO2 produced and the moles of the original gas mixture:
nCO2 = (0.387/RT) * V
ntotal = (0.387/RT) * V
Step 5: Calculate the ratio of moles of CO2 produced to moles of the original gas mixture:
Ratio = nCO2 / ntotal
= [(0.387/RT) * V] / [(0.387/RT) * V]
= 1
Therefore, the ratio of moles of CO2 produced to moles of the original gas mixture is 1:1.