5 Steps to a 5 AP Chemistry

(coco) #1

276  Step 4. Review the Knowledge You Need to Score High


You have 20 minutes to do the following questions. You may use a calculator and the tables
at the back of the book.
The alkane hexane, C 6 H 14 , has a molecular mass of 86.17 g/mol.

a. Like all hydrocarbons, hexane will burn. Write a balanced chemical equation for the com-
plete combustion of hexane. This reaction produces gaseous carbon dioxide and water.
b. The complete combustion of 10.0 g of hexane produces 487 kJ. What is the molar heat
of combustion (ΔH) of hexane?
c. Determine the pressure exerted by the carbon dioxide formed when 5.00 g of hexane are
combusted. Assume the carbon dioxide is dry and stored in a 20.0-L container at 27°C.
d. Under identical conditions, hexane vapor diffuses at one-half the rate of the vapor of
another compound. What is the molar mass of the other compound?
e. Hexane, like most alkanes, may exist in different isomeric forms. The structural for-
mula of one of these isomers is pictured below. Draw the structural formula of any two
other isomers of hexane. Make sure all carbon atoms and hydrogen atoms are shown.

 Answers and Explanations


a. 2C 6 H 14 +19O 2 →12CO 2 +14H 2 O
Give yourself 2 points for the answer shown above, or for the coefficients: 1, 9/2, 6, and 7.
Give yourself 1 point if you have one or more, not all, of the elements balanced.

b. (–487 kJ/10.0 g hexane) (86.17 g hexane/mol hexane) =−4.20× 103 kJ/mol
Give yourself 2 points for the above setup and correct answer (this requires a negative
sign in the answer). If the setup is partially correct, give yourself 1 point.

c. The ideal gas equation should be rearranged to the form P=nRT/V.
n=(5.00 g hexane) (1 mol hexane/86.17 g hexane) (12 mol CO 2 /2 mol hexane)
=0.3481 mol CO 2
This answer has an extra significant figure. The mole ratio should match the one given
in your balanced equation. You will not be penalized again for an incorrectly balanced
equation. You will lose a point if you do not include a hexane-to-CO 2 conversion.
R=0.0821 L atm/mol K
This value is given in your test booklet.
T= 27 °+ 273 =300. K
You will be penalized if your forget to use the Kelvin temperature.
V=20.0 L
P=(0.3481 mol CO 2 ) (0.0821 L atm/mol K) (300. K)/(20.0 L)=0.429 atm.
Give yourself 2 points for the correct setup and answer. Give yourself 1 point if you did
everything correctly, except the mole ratio or the Kelvin conversion.

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