5 Steps to a 5 AP Chemistry

grams of KClO 3 are present. You take the mixture and heat it. The KCIO 3 decomposes
according to the equation:

2 KClO 3 (s) →2 KCl(s) +3 O 2 (g)
The oxygen gas that is formed is collected by displacement of water. It occupies a
volume of 542 mL at 27°C. The atmospheric pressure is 755.0 torr. The vapor pressure of
water at 27°C is 26.7 torr.
First, you need to determine the pressure of just the oxygen gas. It was collected over
water, so the total pressure of 755.0 torr is the sum of the partial pressures of the oxygen
and the water vapor:

The partial pressure of water vapor at 27°C is 26.7 torr, so the partial pressure of the
oxygen can be calculated by:

At this point you have 542 mL of oxygen gas at 728.3 torr and 300. K (27°C +273).
From this data you can use the ideal gas equation to calculate the number of moles of
oxygen gas produced:
PV=nRT
PV/RT=n

You will need to convert the pressure from torr to atm:

(728.3 torr) ×(1 atm/760.0 torr) =0.9583 atm

and express the volume in liters: 542 mL =0.542 L

Now you can substitute the quantities into the ideal gas equation:

(0.9583 atm)(0.542 L)/(0.0821 L · atm/K · mol)(300. K) =n

0.02110 mol O 2 =n

Now you can use the reaction stoichiometry to convert from moles O 2 to moles KClO 3
and then to grams KClO 3 :

Non-Ideal Gases

We have been considering ideal gases, that is, gases that obey the postulates of the Kinetic
Molecular Theory. But remember—a couple of those postulates were on shaky ground. The
volume of the gas molecules was negligible, and there were no attractive forces between the
gas particles. Many times approximations are fine and the ideal gas equation works well.
But it would be nice to have a more accurate model for doing extremely precise work or
when a gas exhibits a relatively large attractive force. In 1873, Johannes van der Waalsintro-
duced a modification of the ideal gas equation that attempted to take into account the volume
and attractive forces of real gases by introducing two constants—aand b—into the ideal gas
equation. Van der Waals realized that the actual volume of the gas is less than the ideal gas
because gas molecules have a finite volume. He also realized that the more moles of gas pres-
ent, the greater the real volume. He compensated for the volume of the gas particles math-
ematically with:
corrected volume =V– nb

(. )

.

0 02110

2

3

122 55

2

3

2

mol O mol KClO^3

mol O

⎛ gKClO

⎝

⎜

⎞

⎠

⎟

11

1 723

3

mol KClO gKClO^3

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟=.

PP POTotalHO 22 =−=755.0 torr 26.7 torr 728.3 torr− =

PPPTotal=+O 2 H O 2 (Dalton’s law)

Gases  111