Question

Ethylene glycol, C2H6O2 (l), is a sweet tasting liquid used as a main component of antifreeze or coolant for cars. The combustion of ethylene glycol is also a key reaction for the formation of polyester.

a. What is the empirical formula for ethylene glycol?

b. The balanced chemical equation for the combustion of ethylene glycol is shown below. Identify the oxidizing agent and the reducing agent for the reaction.

2 C2H6O2 (l) + 5 O2 (g)  4 CO2 (g) + 6 H2O (l)

c. If 150 g of C2H6O2 (l) is reacted with 125.0 L of O2 (g) at STP, what will be the theoretical yield of water produced?

Answer 1

B la) Empirical formular CH3O ru CH₂-oh CM₂ - an Ethylene glycol C₂H6O2 CHO 2444444444AAAAAANNNN 2:612 = 1:3:1 2 (b). C₂H6O2(d) +502 (9) 4 co₂+ 6H2O 22 (1) e oxidizing agent = oxygen molecule - Reducing Agent - Ethylene glycol 3 Emmylene glycol oxidized to co2 ent to that's why it act as Reducing Agent To C₂H6O2 +502 4602 + 6H20 Catste 2 t mot 5 mol 5 Imol (4602 need = 5mol O₂ mot of O2 for 125L -2.42 C₂h602 need = 2.4245 I 125L = 12.10 Imole of Ethylene glycol = 22.4L But we have only had in 150 gram = 5.58mof 2 . 5.50 mol O2 = 150g 62 g/mol molar mass of = 2.42 mol (24602 = 12x2+6+ 16x2 = 24+6+32=62 6 mol Imol = 22.4L SUL

cal) * Now mole of a decide the Amount of ho I Because Oz is a limiting Reagent smol O2 Produce = smol H20 & 1 mol O2 Produce = mole 420 5.50 mol O2 Produce - 6x5.58 mol 420 = 6.696 mol 420 1 mol 420 = 10 gram , H₂O 2 2X1 +16=18 6.696 mol 420 = 6.696x18 = 120.528 gram 33333SVOJU H2O Theoretical yield Of 40 = 20.528 gram. Thank hope this will help you Good luck