Question

In your igloo, you decide to have yourself a Barbeque in your air-tight, 445.0 Liter igloo. Your igloo is pressurized with nothing but 1.250 atm O2(g) at -25.00 °C. Your ethane (C2H6) grill uses a tank which holds 5.40 atm of ethane (C2H6) in a 20.0 Liter tank. a. If all the ethane in your tank were to fill the room, what would be the pressure of ethane? b.You decide to start grilling (read: combusting) and you consume (react) your entire tank of ethane. What is the balanced reaction of ethane, C2H6 (g), combusting with O2(g) - Assume the temperature remains at -25.00 °C. c. What is the pressure in the igloo after the combustion reaction is complete? (assume T= -25.00 °C.) d. What is the mole fraction of O2(g) after the reaction is complete? e. How many grams of CO2 is produced? f. Let's now assume the ethane combustion generated enough to heat to raise the temperature of your insulated igloo to 109.0°C. What is the total pressure?

Answer 1

The main concept is that the volume of air occupied = Volume of igloo - volume of ethane tank =445 -20=425 L

If you are supposed to neglect the volume of ethane tank then where ever I have used the volume to be 425 L,you use the volume as 445 L and put them in the formulas and get the answers.

» Volume of igloo uus liter Volume of ethane tank 20 liter Prusun of Oz (inside ig 100) = 1-25 atm Presun 11 tank = 5.40 atm Temp= -25°C = 248k (a) Volume of O2 = volume of igloo - volume of ethane tank - uus -20 I UPS uter . Molu of ethane 'n PV=nRT -> 5.48 202 nx 0.08 21(248) » n = 5.30 Moles of O2 = 'm' pvz MRT . 1.25 X 425= mx 0. 0821x (248) -> m= 26.09] il all the ethane were to fill the room the p= nRT = 5.3X0-08 21 x 298 = 0.8 6 Hatu 125

2. Pressure of eth ame= 0.884 atm (6) C2 Mo + 7 0 2 2 (02 + 3120 5.3 26.09 (c) la Hot 702 - 2102 r 3120 5.3 26-09 Mules of 02 = 26-09 = 7.45 712 T Moles of Cato = 5:3 | Moles of Cato < Mola of 02 3.51 Cato is limiting -> Ca Ho +70 - (02t 3150 reagent o 26-09-7x3-3 Zx 5.3 3x5.3 o 7.54 1006 15.9 Final male = 10.6+15-9 + 7.54= 34.04 Moly Pүрир) & PX 425= 34.04x10821 x 248 » [P= 1.63 atm]

2 0.20 7.54 34.04 (d) Mele fraction of Or Moles of 02 = 7.54 Total moles 3404 X on = 0.2 (1) weight of CO2 produced a Mass of 102x Melis 24ux 10.6 = 466.4 gram (f) Now temp- 109 °C = 382k Ip van RT - Px 425= 34.04x0-08 21 X 382 - 1P2 2.51 atm.