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

no of moles of 2-methoxy-2-methyl propane = 4.43/88 = 0.0503 moles

PV =nRT
n = PV/RT
= 119.19*1.64*10^-1/ 0.0821*(327.9+273)
= 19.54716/49.33389
= 0.396moles of O2
              C5H12O +   15/2 O2 ----> 5CO2 + 6H2O
Initial       0.0503     0.396       0      0
              0.0503-X   0.396-8X     5X     6X
        
              0.0503-X = 0
              X=     0.0503
              O2 = 0.396-15/2*0.0503 = 0.01875 moles
              Co2 = 5*0.0503 = 0.2515 moles
              H20 = 6*0.0503 = 0.3018 moles
total no of moles = 0.01875+ 0.2515+ 0.3018
                   = 0.57205 moles
P = nRT/V
   = 0.5705*0.0821*600.9K/0.164 = 171.6 atm

mole fraction of CO2 = no of moles of CO2/ total no of moles
PCo2 = mole fraction of CO2* total pressure
     = 0.2515*171.6/0.57205
     = 75.5 atm

PO2 = mole fraction O2 * total pressure
     = 0.01875*171.6/0.57205
     = 5.63 atm

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