Untitled - Kelly Walsh High School

Gases 93
If it was not clear before, it should be clear now, that we still must find moles.
We will find moles from the mass of KClO 3 and the balanced chemical equation.
We need to determine the molar mass of KClO 3 from the atomic weights
of the individual elements (122.55 g/mol). We now add our mole information to
the equation:
B(12.2 g KClO3 )¢
V
1 mol KClO3 3 mol O2 L atm
≤¢
≤R ¢0.0821 ≤(293 K)
122.55 g KClO3 2 mol KClO3 mol K
(737 mmHg)
To complete the problem, we need to add a pressure conversion:
B(12.2 g KClO3 )a
V
1 mol KClO3 ba
122.55 g KClO3 3 mol O2 L atm
b R a0.0821 b(293 K)
2 mol KClO3 mol K
(737 mmHg)
V 3.70420 3.70
¢ 760 mmHg
1 atm ≤
Quick Tip
You should be very careful! when working problems involving gases and one or
more other phases. The gas laws can only give direct information about gases.
This is why there is a mole ratio conversion (from the balanced chemical equation)
in this example to convert from the solid (KClO 3) to the gas (O 2).
The methods shown in this section will apply equally well to nonideal (real)
gases, with van der Waals equation used in place of the ideal gas equation.
However, real gases require the use of van der Waals constants from appropriate
tables.
In this chapter, you learned about the properties of gases. You learned that you
can use the combined gas law, the ideal gas law, or the individual gas laws to calculate
certain gas quantities, such as temperature and pressure. You also
learned that these equations could also be useful in reaction stoichiometry
problems involving gases. You learned the postulates of the Kinetic-Molecular