In: Chemistry
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?
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