Question

Isooctane, C8H18, is the component of gasoline from which the term octane rating derives.

A. Write a balanced equation for the combustion of isooctane to yield CO2 and H2O.

B. Assuming that gasoline is 100% isooctane, the isooctane burns to produce only CO2 and H2O, and that the density of isooctane is 0.792 g/mL, what mass of CO2 in kilograms is produced each year by the annual US gasoline consumption of 4.6 x 10^10 L?

C. What is the volume in L of this CO2 at STP?

D. How many moles of air are necessary for the combustion of 1 mol of isooctane, assuming that air is 21.0% O2 by volume.? What is volume in L of this air at STP?

Answer 1

hu) given Pre 755. 3 toru Po= 23.8 torre So. Po = P - Proo (fon Dalton's lam) POR ² 75503-23.8 PO₂ = 731.5 tour T= 250 298k Solution VEL Step I find moles of O2 PV = nRT n = 1 X 731.5 760 0.0821 X 298 = 0.0394 mal Step 11 find mass of KClO3. 2K410; (5) >> 261(s) + 302 (8) applied POAC Rule. moles of O2 cofficient = moles of kolo, cofficient cofficient mass of KClO3 = 0.894 molar mass X2 3 mass of KClO3 = 0.0394 X 2 X 122.55 3 mass of KClO₂ = 3.229 u ok kelo₂ = 5.22g Thank you

cu) balanced Reaction combution of Tsooctane. 2 Ce Hall) +250,(9) — 1600,(8) + 18 40(8) B) marts of gaxaline volume & Density 4.6X100 [ X 0.7929/2 8.6432 x 1000g = 3.643281010 ne = mass molaumars 114023 = 310894x10tmal Step ☺, applied POAC fou combution of Octane, 2 C9 H10 (l) + 250,(9) > 16 Cox (9) + 18720 () moles of CO2 - moles of Ocatone Cofficient mass of cor - 310894 X107 cofficient molaumass X16 mass of CO2 = 310894 x 10tX 44 X 16 ayolg- 019 mass of (om = 110 23X10*kq

2) at STP, P=latm T= 298k, 8o. PV = nRT latm X V = 11.23x109° X 0.0821x298 unglmal V2 6. 2424X100 L d) again POAC Rule moles of O2 - moles of Cg H18 cofficient officient males of O2 = 1 males of O₂ = 1205 mol so. Vo = nRT. 125 X 0.0821 x 298 25 0. 02 P Voza 30508 22 L so. Vais - Vo2 x 100 21 Vair = 305.822X100 21 Vais = 1456.32 Thank yours