Calculations Based on Chemical Equations
175
„_ (0.0368 moles) (62.4
torr lll f*\ (273 K)
fit\i
\
mole is. /
~T~
=
760 torr
= 0.825 liters O 2 at standard conditions
(c) The volume of O 2 will again be given by the ideal gas law, but the partial
pressure of O 2 must be used (not the combined pressures of O 2 and water
vapor). The partial pressure of O 2 is calculated from Dalton's law, using the
vapor pressure of H 2 O (24 torr) from Table 11-1.
Po, = 730 torr - 24 torr = 706 torr
nRT
(0.0368 moles) 62.4 -
(298 K)
P
706 torr
= 0.969 liters O 2
PROBLEM:
Chlorine is prepared by the reaction
2NaMnO 4 + lONaCl + 8H 2 SO 4 ^ 2MnSO 4 + 6Na 2 SO 4 + 5Cl 2 + 8H 2 O
or
2MnO 4 - + IOC1- + 16H
+ ^ 2Mn
2+ + 5O 2 + 8H 2 O
What weights of (a) pure NaCl and (b) 90.0% pure NaMnO 4 are needed to prepare
500 ml of Cl 2 gas measured dry at 25°C and 730 torr?
SOLUTION:
We follow the three simple steps.
1. From the volume of C1 2 that we are given (to prepare), compute the moles of
C1 2 that are given, using the ideal gas equation.
PV
(730 torr)(0.500 liter)
n = TPF
:
mole K /
= 0.0196 moles C1 2 given
2. From the moles of C1 2 given, compute the number of moles of NaCl and
NaMnO 4 required. The chemical equation shows that 10 moles of NaCl and 2
moles of NaMnO 4 are required for 5 moles of C1 2 , therefore the needed
(
"'IN
"VIT-.V/T" j (0.0196 moles C1 2 ) = 0.0392 moles
10 moles NaCl'
5 moles C1 2
moles of NaMn0 4 =
~
(0.0196 moles C1 2 ) = 0.00784 moles
3. From the moles of NaCl and NaMnO 4 required, express these quantities in
the units specified in the statement of the problem.