Gases Chemistry Worksheet - Chapter 13, An Introduction To Chemistry Page 23
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4913.3 Equation Stoichiometry and Ideal Gases
505
An alternative technique allows us to work Example 13.6 and problems like it
using a single unit analysis setup. This technique, illustrated in Example 13.7, uses the
universal gas constant, R, as a conversion factor.
e
13.7 - Gas Stoichiometry for Conditions Other Than
xample
STP (alternative technique)
The question is the same as in Example 13.6: When 4.0 × 10
7
L of hydrogen gas at
O
19
bjeCtive
503 °C and 155 atm reacts with an excess of nitrogen, what is the maximum volume
of gaseous ammonia that can be formed at 20.6 °C and 1.007 atm?
Solution
We can use R as a conversion factor to convert from volume in liters of H
to moles of
2
H
. We use it in the inverted form, with liters (L) on the bottom, so that liters will be
2
canceled out.
K mol
7
? L NH
= 4.0 × 10
L H
3
2
0.082058 L atm
The universal gas constant (R) is different from other conversion factors in that it
contains four units rather than two. When we use it to convert from liters to moles, its
presence introduces the units of atm and K. We can cancel these units, however, with a
ratio constructed from the temperature and pressure values that we are given for H
:
2
155 atm
K mol
7
? L NH
= 4.0 × 10
L H
3
2
0.082058 L atm
776 K
We now insert a factor to convert from moles of H
to moles of NH
, using the
2
3
coefficients from the balanced equation.
2 mol NH
K mol
155 atm
3
? L NH
= 4.0 × 10
7
L H
3
2
0.082058 L atm
3 mol H
776 K
2
We complete the sequence by inserting R once again, this time to convert moles of
NH
to volume of NH
in liters. We eliminate the K and atm units by using a ratio
3
3
constructed from the temperature and pressure of NH
.
3
2 mol NH
293.8 K
155 atm
0.082058 L
atm
K mol
•
3
= 4.0 × 10
7
? L NH
L H
3
2
1.007 atm
0.082058 L atm
776 K
3 mol H
K
mol
2
•
= 1.6 × 10
9
L NH
3
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