A 8.13 g sample of 2,3,5-trimethylhexane (formula C9H20), is mixed with 26.05 at

Question

A 8.13 g sample of 2,3,5-trimethylhexane (formula C9H20), is mixed with 26.05 atm of O2 in a 2.30 L combustion chamber at 374.49 °C. The combustion reaction to CO2 and H2O is initiated and the vessel is cooled back to 374.49 °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

C9H20 + 14O2 -------> 9CO2 + 10H2O

Moles of C9H20 = mass / MW = 8.13 / 128 = 0.0635

PV = nRT

P = 26.05 atm

R = 0.0821

V = 2.3 L ; T=374.49 +273 = 647.49 K

Substituing in above formula

n = PV/RT = 1.127 moles of O2

1 moles of C9H20 reacts with 14 moles of O2 and gives 9 moles of CO2

0.0635 moles of C9H20 reacts with 0.0635*14 = 0.889 moles of O2 and forms 9*0.0635 = 0.5715 moles of CO2

Moles of CO2 = 0.5715 = n

PV = nRT

P = 13.2 atm

PCO2 = 13.2 atm

Moles of O2 left = 1.127 - 0.889 = 0.238 = n

P =nRT/V = 5.5 atm

PO2 = 5.5 atm

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