A 4.43 g sample of 2-methoxy-2-methylpropane (formula C 5 H 12 O), is mixed with

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 PCO2 (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 PO2 (in atm) in the combustion chamber after the reaction goes to completion?

Explanation / Answer

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

PCO2 = P XCO2

XCO2=0.25/0.575=0.43

PCO2 = 173.1 atm *0.43

PCO2 = 75.26 atm

PCO2 = P XCO2

XO2=0.025/0.575=0.043

PO2 = 173.1 atm *0.43

PO2 = 7.44atm

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