CHEM 103 Module 3 problem set 5 Reviewed Study Guide

1. The combustion of ethanol (C2H5OH) takes place by the following reaction equation. C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g) What is the volume of CO2 gas produced by the combustion of excess ethanol by 23.3 grams of O2 gas at 25oC and 1.25 atm? Your Answer: C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g) number of moles = mass / molar mass Moles of O2 = (23.3 g) / (32.0 g/mol) = 0.728 mole 3 moles of O2 --> 2 moles of CO2 moles of CO2 produced by 0.728 moles of O2 = (2/3) x 1.728 = 0.485 mol Volume of CO2 PV = nRT --> V = nRT/P n = number of moles p = 1.25 atm v = volume of gas t = absolute temp in K (25 + 273) = 298 K r = 0.0821 V = [(0.485 mole) x (0.0821) x (298)] / 1.25 atm V = 9.492 L (MW = 46) (MW = 32) (MW = 44) (MW = 18) C2H5OH (l