Question

Question 9 For a cooking session of 1 hour, 940 mL of propane gas (C3H8, molar mass 44.1 g/mol) are combusted. How much energy is released or required from the combustion of this amount of propane given the following equation and the density of propane (density of C3H8 -0.493 g/mL)? C3H8 (8) + 502 (8) — 3C02 (8) + 4H20 (1) AH = -2.22 x 103 kJ mol-1 § 5 O O OOO 2.33 * 10° K) is absorbed 2.09% 10% kJ is absorbed 2.33 104 kJ is released 1.17 x 104 kJ is released 1.17 x 104 kJ is absorbed

Answer 1

Given reaction C3H8 (g) + 5O2(g) → 3602 (8) + 4H2O(A) AH = -2.22 X103 kJ/moj- Volume of density of propane propane gas - 940 ml gas = 0.493 glme -- mass of C3H₂ = volume x density a 940 ml x 0.493 g/ml - 463.429 CzHg = Given mass .. mol of Mw 4 63.429 4485 Imol = 10:51 mol now, . 1 mol 10.51 mol of Cats on combustion broduces 2.22x103 kJ of heat of C₃H8 on combustion produces 2.22 XIO’x10.5) KJ heat. produced 23.33x103 kJ = 2.33 X204 KJ heat released. so heat C Scanned with CamScanner