Title: Limiting Reagent Exploration
1. Prepare three of the small plastic bottles with 80 mL of vinegar and six of the bottles with 240mL. Put the caps back on to prevent spilling.
2. Fill three balloons (using a funnel) with 5.7 grams (roughly one teaspoon) of baking soda and six balloons with 17.1 grams (roughly three teaspoons) of baking soda. Use different colored balloons to keep track of the different masses of baking soda.
3. Divide students into teams of two or three so that there are nine groups.
4. Move the teams around the room so that there are three teams on the right side, three in the middle, and three on the left. Give the teams on the right side the 80 mL bottles and tell them they are Group A, give the three teams in the middle section three of the 240 mL bottles (Group B), and give the three teams on the left side of the room the other 240 mL bottles (Group C). Have the students record the volume of vinegar inside their bottle.
5. Explain that the balloons (which the students have not yet received) are filled with a certain compound that will react with the vinegar. Have them guess what this compound is. At the front of the room, demonstrate how to put the balloon on top of the bottle without letting any of the baking soda fall out.
6. Demonstrate the reaction for the students. Demonstrate how to measure the circumference of the balloon using a string.
7. Ask the students to predict what will happen when you triple the amount of baking soda or the amount of vinegar.
8. When the students are focused and quiet, pass out the balloons. Give Groups A and B balloons with 17.1 grams of baking soda, and Group C the balloons with 5.7 grams of baking soda. Have them record the mass of baking soda in their balloon and calculate the number of moles of baking soda.
9. Allow the students to react the baking soda and vinegar in the way you demonstrated, and record the circumference of their balloons.
10. Lead a group discussion about why some balloons are inflated more than others.
CH3COOH + NaHCO3 >> CH3COON(aq) + H2O + CO2
Because we are using household vinegar, it can be difficult to calculate the exact number of moles of acetic acid that the experiment uses. While the students should be able to easily calculate the number of moles of baking soda used from the mass, they will not be able to easily convert between milliliters of vinegar and moles of acetic acid. It may be in the teachers best interest to calculate the number of moles of acetic acid ahead of time, and explain to the students how the computation was done. I calculated that about .069 moles of acetic acid are present in 80 mL of household vinegar. I arrived at this value by multiplying the milliliters of household vinegar by the density of acetic acid. Then I multiplied this value by .05, because household vinegar is around 5 percent acetic acid. Explain to the students that this is not an exact calculation, but that it is satisfactory for this experiment.
Many students will predict that the balloon will triple in size when you triple either the amount of baking soda or vinegar. If you use a 1-1 ratio of reactants for the initial demonstration, this will not be the case. If you triple only one of the reactants, there will not be enough of the other to react with it. The amount of product created will be roughly the same as the initial trial, and the balloon circumference will not change by much. It may change a little due to differences in the balloons, or human error in measuring out the reactants. If you don't have an exact 1-1 ratio to begin with, the balloon size will increase for either Group A or C. If you triple the amount of both reactants, the amount of CO2 produced will then increase greatly. Group B should always have the largest balloons. This introduces the idea of limiting reagents in a hands on and visual way.
Questions & Answers:
Question: When you added triple the amount of one reactant but the base amount of the other, what happened? Why? Did you expect this?
Answer: As stated in the explanation, many students will predict that the balloon will triple in size. As a matter of fact, the balloon is a similar size as it was when the teacher did the initial demonstration. This is because the reactant that was not tripled acts as a limiting reagent. It limits the amount of product that is produced.
Question: There were .069 moles of acetic acid present in every 80 mL of household vinegar. Calculate the number of moles of acetic acid used in your trial. Calculate the number of moles of baking soda that were used using the molar mass. Which one was the limiting reagent for your trial? How many grams of CO2 were theoretically produced from your trial?
For Group A, they used 80 mL of household vinegar which is about .069 moles of acetic acid. They had 17.1 grams of baking soda, when divided by the molar mass of sodium bicarbonate (~84 grams/mole) this is equal to ~ 0.2 moles. Thus, the limiting reagent was the acetic acid for this trial. Using this value, you know that 0.2 moles of CO2 should be produced. Multiplying this by the molar mass of CO2 (~44 grams/mole), you calculate that 8.9 grams of CO2 were theoretically produced.
Question: You notice that the ballon size for the trials in which only one reactant was tripled vs. the trial with the base amount of reactants were slightly different, when theoretically they should have been the same (due to the principle of limiting reagents.) What are potential sources of error for this experiment that may have caused this discrepancy?
Answer: It is difficult to know the exact molarity of household vinegar, therefore it is hard to determine exactly how many moles of acetic acid are present. This makes it difficult to achieve a 1-1 ratio of reactants for the initial demonstration. Furthermore, different balloons may have different give and inflate more easily than others. Also, it is possible that not all of the baking soda shakes out of every balloon, which would also change the ratio of reactant to reactant.
Applications to Everyday Life: Explain (don't just list) three instances where this principle can be used to explain other phenomenon.
Ocean Acidification: When CO2 reacts with water in the ocean, carbonic acid is produced in the following equation: CO2 + H2O --> H2CO3. CO2 is the limiting reagent, because there is less CO2 in the ocean than there is water. With large increases of CO2 in the atmosphere from the combustion of fossil fuels, more CO2 is dissolving into the ocean. Thus more carbonic acid is being produced, because the amount of the limiting reagent has been increased. This causes ocean acidification.
Aerobic Respiration: Glucose and oxygen react to form carbon dioxide, water, and energy. We know that the body needs food and air to survive. If there is a limiting amount of either one, your body can't produce as much energy. Depending on the availability of each, either glucose (food) or oxygen (air) can be the limiting reagent.
Rusting of Iron: When water and oxygen come into contact with iron, it causes rust. The more water or oxygen that come into contact with the iron, the more rust that is produced. The equation is as follows: 4Fe + 3O2 + 6H2O --> 4Fe(OH)3. If the amount of exposure to O2 and H2O is restricted, rust will not be produced. The water and oxygen are the limiting reagents in this case, because they determine how much rust is produced.
References: "Baking Soda and Vinegar" from Experimentopia