Question

C3 H₂O (1) +4 0₂ (g) 3 H₂O(g) + 3 CO₂ (9) Consider an experiment where 1,86 mb of the liquid (density = 0.788 g/ml) is combined with the gas. in a 2.50L container at STP. a. What is the percent yield if 2.97g of the carbon - containing product is collected ? b. Using your answer f-1688.7 kJ/mol, calculate the amount of heat released.

Answer 1

C₃H6 Old) + 402 (2) 3 H₂O (g) + 3C02() 1.86 m mass Volume of too Density f (3 HO = 0.788 of me mt Czlt6o (50 1.86 x 0.788 58.00 almol moles I CHO 1.46568 g 0.02617 mol. 58.00 genel B) = 1.465682 molar mass of Ca Hoo volume 2.50 L. at STP, pressure p = I alm. 2 T = 273K Temperature gas constant, R= 0.0821 hahm. mol-lky n - no of moles for From Ideal gas law PV = nRT PV I atm 2.50 L RT 0.0821 Lahn x 273K molk 0.11 15 4 mol.

I mol of Ca Hoo reacts with 4 mol 02 0.02617 mol mol az of Cz HCO u (4 x 0.02617) moles 0.10468 mol O2 fo O₂ remains 0.11154 - 010468 6.86 x 10-3 Hence Cz HcO (2) is the limiting reagent. Imol of C₃ Hoole) produces 3 mol.co2 Νοιώ, 0.02617 mol C3 Holdl 3 x 0.02617 molco2 O 07851 mol molar mass of e02 44 frist of con produced - 44a mass x 0.07851 mol mol 3. u5aung Theoretical field = 31 3, 45400g Actual yield = 2.97 g 2.9 79x 100% 89.976 % 3.45.4uag

Enthalpy of reaction, 4H - - 1688. I KJ mol, the reaction Amount heat releand during 45 0.026 17 b. 078 51 mol KJ Х 4 q = +1688 mol 1% = 137.5/8/445) 64.19 KB