1 / 6

GASES Question 2: 1995 B Free Response

GASES Question 2: 1995 B Free Response. Park, Sherrie Gangluff, per. ¾ AP Chemistry. Propane, C 3 H 8 , is a hydrocarbon that is commonly used as fuel for cooking. A). Write a balanced equation for the complete combustion of propane gas, which yields CO 3 (g) and H 2 O (l).

kendall-tyson
Download Presentation

GASES Question 2: 1995 B Free Response

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. GASESQuestion 2: 1995 BFree Response Park, Sherrie Gangluff, per. ¾ AP Chemistry

  2. Propane, C3H8, is a hydrocarbon that iscommonly used as fuel for cooking. A). Write a balanced equation for the complete combustion of propane gas, which yields CO3 (g) and H2O (l). C3H8(g) + 5O2(g) 3CO2 (g) + 4H2O (l)

  3. B). Calculate the volume of air at 30 C and 1.00 atmosphere that is needed to burn completely 10.0 grams of propane. Assume that air is 21.0 % O2 by volume. • Afterwards, use PV=nRT to find the volume of O2 in liters. • PV = nRT : (1 atm)(x) = (1.14 mol O2)(0.0821 L · atm · K-1 · mol-1 )(303 K) = 28.4 L O2 • First, convert 10.0 g C3H8 into moles. In order to do this, you must divide the given number of grams by the molar mass of C3H8: • (10.0 g C3H8)/(44 g C3H8) = 0.227 mol C3H8 According to the equation C3H8(g) + 5O2(g) 3CO2(g) + 4H2O (l)the mole ratio of O2 to C3H8 is 5:1 As a result.. 0.227 mol C3H8 x (5 mol O2/ 1 mol C3H8) = 1.14 mol O2 Finally, looking back at the question we find the relation of O2 and air being that air is 21% O2 by volume. Therefore: 28.4 L O2 = .21(x) x = 135 L Air With this number, you can then determine the number of moles found in O2 by multiplying 0.227 mol C3H8 with the mole ratio of O2 and C3H8.

  4. C). The heat of combustion of propane is -2,220.1 kJ/mol. Calculate the heat of formation, DHf, of propane given that DHf of H2O (l) = -285.3kJ/mol and DHf of CO2 (g) = -393.5 kJ/mol. Now that we have our equation, we can then substitute in DHf for H2O and CO2 as they are provided for us above. SDHf = [3 DHf CO2(g) + 4 DHf H2O (l) ] – [DHf X + 0] = [3(-393.5) + 4(-285.3)] – [X + 0] X = DHf of C3H8 = Now that you know this you can complete the problem by first analyzing the balanced equation that was determined in question A. C3H8(g) + 5O2(g) 3CO2(g) + 4H2O (l) The result of this equations leads to.. SDHf = [3 DHf CO2 (g) + 4 DHf H2O(l)] – [DHf X + 0] The term X is the heat of formation of propane, or what we are trying to find. To calculate the heat of formation, you must use the equation below: Remember! All pure elements are equal to zero. ΔHf = SDHf products - SDHf reactants -101.7 kJ/mol

  5. D). Assuming that all of the heat evolved in burning 30.0 grams of propane is transferred to 8.00 kilograms of water (specific heat = 4.18 J/gK), calculate the increase in temperature of water. Solving this problem requires you to use the equation: q = (m)(Cp)(∆T) You already know the value of m (or mass – which is equal to 8.00 kilograms H2O) and Cp (or specific heat capacity – equal to 4.18 J/gk for water). However, what we need to find is the temperature differential as well as Q (or the heat energy – which is 1514 kJ when using 30.0 g C3H8 in the equation above). Since you have all your values, substitute them into the equation: q = (m)(Cp)(∆T) = 1514 kJ = (8.00 kg)(4.18 J/g.K)(∆T) =∆T = 45.3˚ 

  6. The End!!!

More Related