Question

A 3.48 g sample of 1,5,5-trimethylcyclohexene (formula C9H16), is mixed with 45.88 atm of O2 in a 3.77×10-1 L combustion chamber at 209.48 °C. The combustion reaction to CO2 and H2O is initiated and the vessel is cooled back to 209.48 °C. Report all answers to two decimal places in standard notation (i.e. 1.23 atm).

What is PH2O (in atm) in the combustion chamber assuming the reaction goes to completion? Assume all water is water vapour and that the volume of the system does not change.

What is PO2 (in atm) in the combustion chamber after the reaction goes to completion?

Answer 1

First to all, you need to write, and then, balance the overall reaction:

C₉H₁₆ + 12O₂ --------> 9CO₂ + 8H₂O

Once you balance, it's time to calculate the number of moles of C9H16 and O2:

moles of C₉H₁₆ : 3.48 g / (9x12 + 16*1) g/mol = 0.0281 moles

moles of O₂ = PV / RT = 45.88 atm * 0.377 L / 0.082 L atm/mol K * (209.48+273) K = 0.4372 moles

Now let's calculate the limitant reactant:

1 mol C₉H_{16 --------->} 12 mol O₂

0.0281 moles ---------> X

X = 0.0281 * 12 = 0.3372 moles of O₂ and we have 0.4372 moles, this means that the limitant reactant is C₉H₁₆:

C₉H₁₆ + 12O₂ --------> 9CO₂ + 8H₂O

i. 0.0281 0.4372 0 0

c. -0.0281 (0.4372-0.3372) +0.0281 +0.0281

e. 0 0.1 0.0281 0.0281

This means that the moles of H₂O is 0.0281 moles * 8 = 0.2248 moles

now the pressure:

P = nRT / V

P = 0.2248 * 0.082 * 482.48 / 0.377

PH₂O = 23.59 atm

For the O₂:

P = 0.1 * 0.082 * 482.48 / 0.377

PO₂ = 10.49 atm

Hope this helps