Review the following concepts if necessary:

- Kinetic Molecular Theory for AP Chemistry
- Gas Law Relationships for AP Chemistry
- Boyle's, Charles's, Gay-Lusaac's and Avogadro's Gas Law for AP Chemistry

### Questions

**First Free-Response Question**

You have 20 minutes to do the following question. You may use a calculator and the tables at the back of the book.

A hydrogen gas sample is collected over water. The volume of the sample was 190.0 mL at 26°C, and the pressure in the room was 754 mm Hg. The vapor pressure of water at 26°C is 25.2 mm Hg.

- Calculate the number of moles of hydrogen in the sample.
- Calculate how many molecules of water vapor are present in the sample.
- Determine the density (in g/L) of the gas mixture.
- Determine the mole fraction of water.

**Second Free-Response Question**

You have 20 minutes to do the following questions. You may not use a calculator. You may use the tables in the back of the book.

A sample containing 2/3 mole of potassium chlorate, KClO_{3}, is heated until it decomposes to potassium chloride and oxygen gas. The oxygen is collected in an inverted bottle through the displacement of water. Answer the following questions using this information.

- Write a balanced chemical equation for the reaction.
- How many moles of oxygen gas are produced?
- The temperature and pressure of the sample are adjusted to STP. The volume of the sample is found to be slightly greater than 22.4 liters. Explain.
- An excess of sulfur is burned in the oxygen. Write a balanced chemical equation and calculate the number of moles of gas formed.
- After the sulfur had completely reacted, a sample of the residual water was removed from the bottle and found to be acidic. Explain.

### Answers and Explanations

**First Free-Response Question**

*P*_{total}= 754 mm Hg*P*_{hydrogen}= 754 – 25.2 = 729 mm Hg*P*_{water}= 25.2 mm Hg/760 mm Hg = 0.0332 atm.*T*and*V*are the same as in part**a**. Avogadro's number = 6.022 × 10^{23}molecules mol^{–1}=*N*- (0.00742 mol H
^{2}) × (2.016 g H^{2}/mol H^{2}) = 0.0150 g H^{2} - 0.00742 mol H
^{2}and 2.57 × 10^{–4}mol H2O

*P* = 729 mm Hg/760 mm Hg = 0.959 atm

*V* = 190.0 mL = 0.1900 L *T* = 26°C = 299 K

*R* = 0.0821 L atm mol^{–1} K^{–1}

*n* = *PV/RT*

= (0.959 atm × 0.1900 L)/(0.0821 L atm mol^{–1} K^{–1} × 299 K)

= 0.00742 mol H_{2}

Give yourself 1 point for the correct answer (no deduction for rounding differently). You must include *all* parts of the calculation (including "=").

Give yourself 1 point for the correct equation, or for any other correct calculation. Do not give yourself more than 2 points total for this part.

*n* = *PV/RT*

= (0.0332 atm × 0.1900 L)/(0.0821 L atm mol^{–1} K^{–1} × 299 K)

= 2.57 × 10^{–4} mol H_{2}O)

molecules = (2.57 × 10^{–4} mol) × (6.022 × 10^{23} molecules mol^{–1})

= 1.55 × 10^{20} molecules

Give yourself 1 point for the correct answer (no deduction for rounding differently). You must include *all* parts of the calculation (including "=").

Give yourself 1 point for the correct equation, or for any other correct calculation.

Do not give yourself more than 2 points total for this part.

(2.57 × 10^{–4} mol H^{2}O) × (18.02 g H^{2}O/mol H^{2}O) = 0.00463 g H^{2}O

total mass = 0.0150 g H^{2} + 0.00463 g H^{2}O = 0.0196 g

density = 0.0196 g/0.1900 L = 0.103 g/L

Give yourself 1 point for the correct answer (no deduction for rounding differently). You must include *all* parts of the calculation (including "=").

Give yourself 1 point for the correct equation, or for any other correct calculation.

Do not give yourself more than 2 points total for this part.

total moles = (0.00742 + 2.57 × 10^{–4}) = 0.00768 mol

Mole fraction H^{2}O = (2.57 × 10^{–4} mol H^{2}O)/(0.00768 mol)

- = 0.0335

*all* parts of the calculation (including "=").

Give yourself 1 point for the correct equation, or for any other correct calculation.

Do not give yourself more than 2 points total for this part.

Total score = sum of parts **a**–**d**. If any of the final answers has the incorrect number of significant figures, subtract one point.

**Second Free-Response Question**

- 2 KClO
_{3}(s) → 2 KCl(s) + 3 O_{2}(g) - (2/3 mole KClO
_{3}) (3 moles O_{2}/2 moles KClO_{3}) = 1 mole O_{2} - At STP the volume of 1 mole of O
_{2}should be 22.4 liters. The volume is greater because oxygen was not the only gas in the sample. Water vapor was present. The presence of the additional gas leads to a larger volume. - There are two acceptable equations; either will get you 1 point. You do not need both equations.
- S(s) + O

_{2}(g)→ SO_{2}(g)- 2 S(s) + 3 O

_{2}(g) → 2 SO_{3}(g) - Either of the possible sulfur oxides will dissolve in water to produce an acid. This will get you 1 point, as will a similar comment and either of the following equations:
- SO

_{2}+ H_{2}O → H_{2}SO_{3}- SO

_{3}+ H_{2}O → H_{2}SO_{4}

You get 1 point if you have the above equation.

You get 1 point for the correct answer and 1 point for the work. You can get these points if you correctly use information from an incorrect equation in part **a**.

You get 1 point for discussing STP and 22.4 liters, and 1 point for discussing the presence of water vapor.

If you chose the first equation, the moles of gas produced would be:

- (1 mole O

_{2}) (1 mole SO

_{2}/1 mole O

_{2}) = 1 mole SO

_{2}

If you chose the second equation, the moles of gas produced would be:

- (1 mole O

_{2}) (2 mole SO

_{3}/3 mole O

_{2}) = 2/3 mole SO

_{2}

You get 1 point for either of these solutions. You will also get 1 point if you used an incorrect number of moles of O_{2} from a wrong answer for part **b**.

Total your points for the different parts. There are 8 points possible.

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