A 8.90-g sample of ethane, C2H6, is mixed with 18.8 atm of O2 (an excess) in a 1.50 L combustion chamber at 130.0°C. The combustion reaction to CO2 and H2O is initiated and the vessel is cooled back to 130.0°C. What is the final pressure in the combustion chamber assuming the reaction goes to completion? Express your answer in scientific notation.

Write the equation and balance it.

Convert 8.9 g C2H6 to mols.
Convert 18.8 atm oxygen @ 130.0 C to mols using PV=nRT.
Determine the limiting reagent. I THINK it is oxygen which means all of the oxygen will be used, there will be some ethane unreacted. Check me out on this.
Determine mols H2O and mols CO2 formed and mols C2H6 unreacted. Add those three together, use PV=nRT to determine pressure of those three materials, Then that number is to be subtracted from the combined pressure of oxygen AND C2H6 at the beginning. (Obviously you will need to use PV = nRT to determine the pressure of C2H6 initially.).
Check my thinking. Post your work if you get stuck.

To find the final pressure in the combustion chamber, we need to calculate the moles of carbon dioxide (CO2) and water (H2O) produced in the reaction.

First, let's calculate the moles of ethane (C2H6) in the sample:
Molar mass of ethane (C2H6) = (2 * Molar mass of carbon) + (6 * Molar mass of hydrogen)
= (2 * 12.01 g/mol) + (6 * 1.01 g/mol)
= 24.02 g/mol + 6.06 g/mol
= 30.08 g/mol

Moles of ethane = Mass of ethane / Molar mass of ethane
= 8.90 g / 30.08 g/mol
≈ 0.296 mol

Since ethane is completely combusted, it will produce the same number of moles of carbon dioxide (CO2) and water (H2O). Therefore, the moles of CO2 and H2O produced are also 0.296 mol each.

Now, let's calculate the total number of moles of gas in the combustion chamber:
Total moles of gas = Moles of ethane + Moles of oxygen
= 0.296 mol + 0 (since oxygen is in excess)
= 0.296 mol

According to the ideal gas law, 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 in Kelvin.

Since the temperature is constant and the volume is fixed at 1.50 L, we can rearrange the equation to solve for the pressure:

P = (nRT) / V

Substituting the values, we get:
P = (0.296 mol * 0.0821 L·atm/mol·K * 403.15 K) / 1.50 L

Calculating this expression, we find:
P ≈ 7.95 atm

Therefore, the final pressure in the combustion chamber, assuming the reaction goes to completion, is approximately 7.95 atm.