Question

Please show steps clearly.

The question is:

9.45 g of liquid hexane (C6H14) is introduced into a 10.0 L vessel containing 13.15 atm of oxygen gas at 21 degrees celsius and ignited, yielding carbon dioxide and water. If the vessel is then cooled to - 10 degrees celsius, what will be the gas pressure inside the vessel?

Answer 1

. Molar mass of henome = 86:18 g/mol Moles = mass Molar mass Moles of hexane = 9.459. 0.11 mol 86. 18 g/mol for 9; N= 1.002., P =13.15 atm Temperature -21° a T=273721 5 294k R = 0.082 Labm/molk Ideal gas law, PV=ART moles of 02, n= PA = (13.15 atm) 18.002). (0.082 Lab) ( 294k) n= 545 MOI Moles of o 2545 mol Moles of hexane .0.11 mo) Reaction is: 266 Hii + 190, - 12202 + 1440 2mol 19mol 12moi lumol (Stoichioroch - coeft) 2mol ca tu needs 19 mot os = 1 mol Coty needs 19 mol og => 0.11 mol CG HW needs 1920.11) = 1045 mol But we only had on sous mol % ' which is more than beanized So Limiting reactant = 0 Go Hui excess reactant - Gotta 02

Co Hinduces 12 * coz 2 mol Os produces 12 moi 3 Imol en produces 6 mol CO2 > 01 mol a Cory produces 620-11) 50-66 mol Moes of con produced =0.66 mot 2 mol COHlu produces 14 mol to Imol C6Huy produces mol H2O oll mol Hiu produces 18011)=0.72 mol H20 moles of Hio produced comm! s Motes of Od present – 3.45 mol Moles of a required for reaction = 1.045 mol Moles of Os left after reaction - 4.405 mol Motes present in vessel = 0.66707744405 na 5.835 mol e volume remaine same as vessel is same T= -10°C - T=273-10 = 263k. on v= 10:02. Pe basats, R=0.882 Lotm Pressure P = ART = (5:835ml) (0108 lat (253) 1002 p 12.58 alm. So, Pressure inside the vessel = 12.6 abm)