Transcript GASES Question 2: 1995 B Free Response

GASES
Question 2: 1995 B
Free Response
Park, Sherrie
Gangluff, per. ¾
AP Chemistry
Propane, C3H8, 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 CO3 (g)
and H2O (l).
C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (l)
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
findinto
the volume
oforder
O2 in
•First, convert
10.0 g to
C3H8
moles. In
liters.
to do this, you must divide the given number of
theatm)(x)
molar mass
of mol
C3H8:
•PV grams
= nRTby
: (1
= (1.14
O2)(0.0821 L · atm ·
K-1 · mol-1
K) = 28.4g LC3H8)
O2 = 0.227 mol
•(10.0)(303
g C3H8)/(44
According
to the equation C3H8 (g) + 5O2 (g)  3CO2 (g) +
C3H8
Finally,
looking back at the question we find the relation of
4H2OWith
(l) the mole ratio of O2 to C3H8 is 5:1
this
number,
you
can
then
determine
the
O2 and air being that air is 21% O2 by volume. Therefore:
number of moles found in O2 by multiplying
As a result..
28.4 L
O2C3H8
= .21(x)
0.227
mol
with the mole ratio of O2
0.227and
molC3H8.
C3H8 x (5 mol O2/ 1 mol C3H8) = 1.14 mol O2
x = 135 L Air
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
have
ourthis
equation,
can thenthe
substitute
Nowthat
thatwe
you
know
you canwe
complete
problem
inbyDH
f for
H2O andthe
CObalanced
2 as they equation
are provided
first
analyzing
that for
wasus
To calculate
heat of
above.
determined
in the
question
A.formation, you must use
the equation below:
S DHf C
=3[3
f CO
DHf 2H(g)2O+ (l)4H
] –2O[DH
H8DH
(g) +
5O22 (g)
(g) +
43CO
(l) f X + 0]
ΔHf = S DHf products - S DHf reactants
= [3(-393.5)
+ 4(-285.3)]
– [X
The result
of this equations
leads
to..+ 0]
Remember! All pure elements are equal to zero.
DH
f of
C32H(g)8 =+ 4-101.7
S DHX
f ==[3
DH
f CO
DHf H2kJ/mol
O(l)] – [DHf X + 0]
The term X is the heat of formation of propane, or what
we are trying to find.
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:
Since you have all your values, substitute them
q =into
(m)(Cp)(∆T)
the equation:
Youq already
know the value of m (or mass – which is
= (m)(Cp)(∆T)
equal to 8.00 kilograms H2O) and Cp (or specific heat
= 1514 kJ = (8.00 kg)(4.18 J/g.K)(∆T)
capacity – equal to 4.18 J/gk for water). However,
what
we =
need
to find is the temperature differential as
= ∆T
45.3˚
well as Q (or the heat energy – which is 1514 kJ when
using 30.0 g C3H8 in the equation above).
The End!!!