Calculate the ratio of moles of CO2 produced to moles of the original gas mixture?

A mixture of methane (CH4) and ethane (C2H6) is stored in a container at 294 mmHg. The gases are burned in air to form CO2 and H2O. If the pressure of CO2 is 351 mmHg measures at the same temperature and volume as the original mixture, calculate the ratio of moles of CO2 produced to moles of the original gas mixture?
Can someone please help me solve this question. I've been trying to figure this out forever. Please show me step by step on how to solve this. I know we have to use the ideal gas law expression for both carbon dioxide and methane and ethane gas mixture. Any help is greatly appreciated!
Thanks :)

well, the moles will be proportional to pressure.

Balance the equations:
CH4+2O2>>CO2 + 2H20 **1
C2H6+7.5 O2>> 2CO2 + 3H2O **2

so in methane, one gets the same amount of CO2 as CH4
In the ethane, one gets twice as much.

So,of the final pressure, the difference is one of the CO2 from the ethane, which is the same amount as the 1/2 the original ethane.

Pressure(351-294)mmHg. Then the pressure due to the other 1/2 of ethane reaction is the same.

Now we can examine the mole ratio:
partial pressure CO2 due to ethane= 2(351-294)=114mmHg
partial pressure CO2 due to methane=357-114= 243mmHg

so moles methane/ethane= 243/114
check my thinking.

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.