General Chemistry Exam 3 Review – Step-by-Step Study Guidance

Q1. If 10 g of sodium bicarbonate (NaHCO3) is combined with an excess of hydrochloric acid (HCl), how many liters of carbon dioxide (CO2) gas are produced at 1.0 atm and 20 °C?

Background

Topic: Stoichiometry and Gas Laws

This question tests your ability to use stoichiometry to determine the amount of gas produced in a chemical reaction, and then apply the ideal gas law to find the volume of gas at given conditions.

Key Terms and Formulas:

• Molar mass of NaHCO3: 84.01 g/mol

• Ideal Gas Law: $PV = nRT$

• R (gas constant): $0.0821\ \mathrm{L\cdot atm\cdot mol^{-1}\cdot K^{-1}}$

• Temperature must be in Kelvin: $T(\mathrm{K}) = T(\degree C) + 273.15$

Step-by-Step Guidance

1. Write the balanced chemical equation for the reaction:

   $\mathrm{NaHCO_3\ (s) + HCl\ (aq) \rightarrow CO_2\ (g) + H_2O\ (l) + NaCl\ (aq)}$

2. Calculate the number of moles of NaHCO3 used:

   $n_{\mathrm{NaHCO_3}} = \frac{10\ \mathrm{g}}{84.01\ \mathrm{g/mol}}$

3. From the balanced equation, determine the mole ratio between NaHCO3 and CO2. For every 1 mole of NaHCO3, 1 mole of CO2 is produced.

4. Calculate the moles of CO2 produced (it will be the same as the moles of NaHCO3 since the ratio is 1:1).

5. Convert the temperature to Kelvin:

   $T = 20\ \degree C + 273.15$

6. Set up the ideal gas law to solve for volume:

   $V = \frac{nRT}{P}$

Try solving on your own before revealing the answer!

Final Answer: 2.9 L

Using $n = 0.119$ mol, $R = 0.0821$, $T = 293.15$ K, and $P = 1.0$ atm:

$V = \frac{(0.119)(0.0821)(293.15)}{1.0} \approx 2.9$ L

The calculation shows the correct use of stoichiometry and the ideal gas law to find the volume of CO2 produced.

Q2. What are the three principles of kinetic molecular theory?

Background

Topic: Kinetic Molecular Theory of Gases

This question tests your understanding of the basic postulates that describe the behavior of ideal gases at the molecular level.

Key Terms:

• Kinetic Molecular Theory (KMT)

• Gas particles, velocity, elastic collisions

Step-by-Step Guidance

1. Recall that the first principle relates to the size of gas particles compared to the volume of the container.

2. The second principle involves the relationship between particle velocity and temperature (in Kelvin).

3. The third principle addresses the nature of collisions between particles and with the container walls.

Try stating the three principles in your own words before checking the answer!

Final Answer:

1. Gas particles are very small compared to the distances between them (volume of particles is negligible).

2. Average kinetic energy (and thus velocity) of particles is proportional to temperature in Kelvin.

3. Collisions between particles and with the container walls are perfectly elastic (no energy lost).

Q3. What is pressure?

Background

Topic: Gas Properties

This question tests your understanding of the physical meaning and units of pressure.

Key Terms:

• Pressure ($P$): Force per unit area

• Common units: atm, Pa, psi

Step-by-Step Guidance

1. Recall the definition of pressure in terms of force and area:

   $P = \frac{F}{A}$

2. Think about how this applies to gases: gas particles exert force when they collide with the walls of their container.

3. Remember common units: 1 atm = 101,325 Pa = 14.7 psi.

Try defining pressure in your own words before checking the answer!

Final Answer:

Pressure is the force exerted per unit area, often measured in atmospheres (atm), Pascals (Pa), or pounds per square inch (psi).

Q4. Rationalize Boyle’s Law, Charles’s Law, Avogadro’s Law, and Dalton’s Law according to kinetic molecular theory.

Background

Topic: Gas Laws and Kinetic Molecular Theory

This question asks you to connect the macroscopic gas laws to the microscopic behavior of gas particles as described by kinetic molecular theory.

Key Terms and Laws:

• Boyle’s Law: $P \propto \frac{1}{V}$ (at constant T, n)

• Charles’s Law: $V \propto T$ (at constant P, n)

• Avogadro’s Law: $V \propto n$ (at constant P, T)

• Dalton’s Law: $P_{\text{total}} = \sum P_i$

Step-by-Step Guidance

1. For Boyle’s Law, consider how decreasing volume increases the frequency of particle collisions with the container walls, thus increasing pressure.

2. For Charles’s Law, think about how increasing temperature increases particle velocity, which increases pressure unless the volume also increases.

3. For Avogadro’s Law, adding more particles (at constant T and P) requires a larger volume to keep pressure constant.

4. For Dalton’s Law, each gas acts independently, so total pressure is the sum of the pressures each gas would exert alone.

Try explaining each law in terms of particle behavior before checking the answer!

Final Answer:

• Boyle’s Law: Decreasing volume increases collision frequency, raising pressure.

• Charles’s Law: Increasing temperature increases particle speed, requiring more volume to keep pressure constant.

• Avogadro’s Law: More particles mean more collisions, so volume must increase to keep pressure constant.

• Dalton’s Law: Each gas’s particles act independently, so their pressures add up.

Q5. If NH3 and HCl are each released into still air, how far will the NH3 have effused when the HCl has diffused 1.00 meter?

Background

Topic: Graham’s Law of Effusion/Diffusion

This question tests your ability to use Graham’s Law to compare the rates of diffusion (or effusion) of two gases based on their molar masses.

Key Formula:

$\frac{\text{Rate}_{\mathrm{NH_3}}}{\text{Rate}_{\mathrm{HCl}}} = \sqrt{\frac{M_{\mathrm{HCl}}}{M_{\mathrm{NH_3}}}}$

• $M_{\mathrm{HCl}} = 36.461\ \mathrm{g/mol}$

• $M_{\mathrm{NH_3}} = 17.031\ \mathrm{g/mol}$

Step-by-Step Guidance

1. Write Graham’s Law for the rates of diffusion:

   $\frac{\text{Rate}_{\mathrm{NH_3}}}{\text{Rate}_{\mathrm{HCl}}} = \sqrt{\frac{M_{\mathrm{HCl}}}{M_{\mathrm{NH_3}}}}$

2. Plug in the molar masses for HCl and NH3:

   $\sqrt{\frac{36.461}{17.031}}$

3. Calculate the ratio to find how much farther NH3 travels compared to HCl in the same time.

4. Multiply the distance HCl travels (1.00 m) by this ratio to find the distance NH3 travels.

Try setting up the calculation before checking the answer!

Final Answer: 1.46 meters

NH3 travels 1.46 times farther than HCl, so when HCl has diffused 1.00 meter, NH3 has diffused 1.46 meters.