Question

4) Refer to the following information for parts a -e: Ethylene glycol, C;H.O. (b. 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. Identity the oxidizing agent and the reducing agent for the reaction. | 2 C2H40 (0 + 5 05 (g) + 4 COị (g) + 6 H30 (1) c. If 150 g of CH.0 () is reacted with 125.0 L of O. (g) at STP, what will be the theoretical yield of water produced? d. If the combustion of ethylene glycol produces 125.0 L CO. () at a temperature of 26 °C and a pressure of 800 torr, how many moles of CO2 9) were produced from the reaction Page 4 of 8

Answer 1

ra) empirical formula is simplest Coefficient one cb rą is oxidizing agent ! 5 Me®z is reducing agent moles S02E - 1509 62.07 g/mey = 2.4166 mol moles q = ns pu - 1 atm X 125102 10.082057 Lamm x 298K) = 5/12 moi Q is limiting Reactant here, so moles to produced 1112 mol = 6.1344 mol Theoretical yield a loater = moles HOX melar Mass 2 HO = 6.1344 mol x 18.01528 g/mol = 110.51 Theoretical Yield ? n = put P= 800 tore=1.05263 atm V = 125.01 T= 26°C = 26 +273 = 299* R= D.082057 Liatmo k 0.05263 atm ( 125.0L) (0.082057 L.atm x 299K) malik n = 5.363 mol moles cag1 produced = 5.363 mol