Question

A 4.43 g sample of 2-methoxy-2-methylpropane (formula C₅H₁₂O), is mixed with 119.19 atm of O₂ in a 1.64×10^-1 L combustion chamber at 327.9 °C. The combustion reaction to CO₂ and H₂O is initiated and the vessel is cooled back to 327.9 °C. Report all answers to two decimal places in standard notation (i.e. 1.23 atm).



1. What is P_CO2 (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.

2. What is P_O2 (in atm) in the combustion chamber after the reaction goes to completion?

Answer 1

MM(C₅H₁₂O) = 88 g/mol
4.43g / 88 g/mol = 0.050 mol

PV/RT = 119.19atm*1.64 ×10^-1 L L / 0.082*(327.9+273)K

19.56/49.3 = 0.40 mol of O2

C₅H₁₂O + 15/2 O2 -------> 5 CO2 + 6 H2O(g at 215°C)
0.050 ____0.40_________0_______0__
0.050__0.40 -15/2X_____5X_____6X_

x= 0.050
=>O2 = 0.025 mol
CO2 = 0.25mol
H2O = 0.3mol

Tot = 0.575 mol

P = nRT/V = 0.575 *0.082*(327.9+273)/ 1.64×10^-1 L

P = 173.1 atm this total pressure

P_CO2 = P X_CO2

X_CO2=0.25/0.575=0.43

P_CO2 = 173.1 atm *0.43

P_CO2 = 75.26 atm

P_CO2 = P X_CO2

X_O2=0.025/0.575=0.043

P_O2 = 173.1 atm *0.43

P_O2 = 7.44atm