CK-12-Chemistry Intermediate

http://www.ck12.org Chapter 14. The Behavior of Gases

Gas Stoichiometry

In the chapterStoichiometry, you learned how to use molar volume to solve stoichiometry problems for chemical
reactions involving one or more gases at STP. Now, we can use the ideal gas law to expand our treatment of chemical
reactions to solve stoichiometry problems for reactions that occur at any temperature and pressure.

Sample Problem 14.7: Gas Stoichiometry and the Ideal Gas Law

What volume of carbon dioxide is produced by the combustion of 25.21 g of ethanol (C 2 H 5 OH) at 54°C and 728
mmHg? Assume the gas is ideal.

Before using the ideal gas law, it is necessary to write and balance the chemical equation. Recall that in combustion
reactions, the given substance reacts with O 2 to form CO 2 and H 2 O. Here is the balanced equation for the combustion
of ethanol:

C 2 H 5 OH(l)+3O 2 (g)→2CO 2 (g)+3H 2 O(l)

Step 1: List the known quantities and plan the problem.

Known

mass of C 2 H 5 OH = 25.21 g

molar mass of C 2 H 5 OH = 46.08 g/mol

P = 728 mmHg

T = 54°C = 327 K

Unknown

volume of CO 2 =? L

The number of moles of carbon dioxide gas is first calculated by stoichiometry. Then, the ideal gas law is used to
calculate the volume of CO 2 produced.

Step 2: Solve.

25 .21 g C 2 H 5 OH×

1 mol C 2 H 5 OH 46 .08 g C 2 H 5 OH

×

2 mol CO 2 1 mol C 2 H 5 OH = 1 .094 mol CO 2

The moles of ethanol (n) is now substituted into PV=nRT to solve for the volume.

V=

nRT P

=

1 .094 mol× 62 .36 L·mmHg/K·mol×327 K 728 mmHg

= 30 .6 L

Step 3: Think about your result.

The mass of ethanol is slightly more than one half mole, meaning that the mole ratio results in slightly more than
one mole of carbon dioxide being produced. Because of the elevated temperature and reduced pressure compared to
STP, the resulting volume is larger than 22.4 L.

CK-12-Chemistry Intermediate

×

V=

=

= 30 .6 L

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