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GASES Question 2: 1995 B Free Response Park, Sherrie Gangluff, per. ¾ AP Chemistry.

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Presentation on theme: "GASES Question 2: 1995 B Free Response Park, Sherrie Gangluff, per. ¾ AP Chemistry."— Presentation transcript:

1 GASES Question 2: 1995 B Free Response Park, Sherrie Gangluff, per. ¾ AP Chemistry

2 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). C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (l)

3 Finally, looking back at the question we find the relation of O 2 and air being that air is 21% O 2 by volume. Therefore: 28.4 L O 2 =.21(x) x = 135 L Air According to the equation C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (l) the mole ratio of O 2 to C 3 H 8 is 5:1 As a result mol C 3 H 8 x (5 mol O 2 / 1 mol C 3 H 8 ) = 1.14 mol O 2 With this number, you can then determine the number of moles found in O 2 by multiplying mol C 3 H 8 with the mole ratio of O 2 and C 3 H 8. Afterwards, use PV=nRT to find the volume of O 2 in liters. PV = nRT : (1 atm)(x) = (1.14 mol O 2 )( L · atm · K-1 · mol-1 )(303 K) = 28.4 L O 2 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 % O 2 by volume. First, convert 10.0 g C 3 H 8 into moles. In order to do this, you must divide the given number of grams by the molar mass of C 3 H 8 : (10.0 g C 3 H 8 )/(44 g C 3 H 8 ) = mol C 3 H 8

4 To calculate the heat of formation, you must use the equation below: Remember! All pure elements are equal to zero. C). The heat of combustion of propane is -2,220.1 kJ/mol. Calculate the heat of formation, D H f, of propane given that D H f of H 2 O (l) = kJ/mol and D H f of CO 2 (g) = kJ/mol. ΔHf = S DHf products - S DHf reactants Now that you know this you can complete the problem by first analyzing the balanced equation that was determined in question A. C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 O (l) The result of this equations leads to.. S D H f = [3 D H f CO 2 (g) + 4 D H f H 2 O (l) ] – [ D H f X + 0] The term X is the heat of formation of propane, or what we are trying to find. Now that we have our equation, we can then substitute in D H f for H 2 O and CO 2 as they are provided for us above. S D H f = [3 D H f CO 2 (g) + 4 D H f H 2 O (l) ] – [ D H f X + 0] = [3(-393.5) + 4(-285.3)] – [X + 0] X = D H f of C 3 H 8 = 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 H 2 O) 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 C 3 H 8 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!!!


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