Limiting Reagent Lab: The reaction between vinegar and baking soda. Goal: During this lab students will gain a quantitative understanding of limiting reagents. Introduction: See textbook sections on stoichiometry and limiting reagent. Safety: Safety goggles should be wom at all times. Students should hold the balloons on the test tubes tightly while the reaction takes place. Procedure: 1) Weigh the following six amounts of baking soda (sodium bicarbonate, NaHCO3): 0.20 g. 0.40 g. 0.50 g, 0.70 g, 1.00 g, and 1.70 g (use a weighing paper to do it). 2) Label the balloons 1-6. Put the six different masses of baking soda into six balloons using a small funnel. Make sure the baking soda goes to the bottom of the balloon. 3) Using the graduated cylinder accurately measure and transfer 10.0 mL vinegar (5% acetic acid, HC,H,O2) into each of the 6 test tubes. 4) Attach the filled balloons to the mouth of the test tubes. Make sure that the contents of the balloon and test tube are not mixed. 5) Before mixing the contests of the balloons and test tubes, make a prediction about which combination will produce the greatest amount of carbon dioxide gas (hypothesis). 6) After the balloons are securely attached to the test tubes, lift a balloon on one of the test tube so that the content of the balloon mix with the test tube content. All six test tubes should be reacted simultaneously, if possible. GET HELP! When the reactions are done, record your results in the table provided. Final Report: DATA TABLE see results and analysis section to help fill out table Test Tube + Mass of Moles of Volume NaHCO, NaHCO, of acid, (8) (ml.), Moles of acetic acid Experimental Reactant Limiting mole ratio in excess reactant NaHCO, acetic acid Moles of CO, formed 1 0.20 g 10.0 2 0.40 g 10.0 3 0.50 g 10.0 4 0.70 g 10.0 5 1.00 g 10.0 6 1.70 g 10.0 Results and analysis (Post Lab): (You must show all your work in your report). 1) Rank the test tubes in order of the least to the greatest amount of carbon dioxide gas (CO2) produced. 2) Write the balanced molecular equation for the reaction that took place. 3) What is the identity of the gas that inflated the balloons? 4) What is the theoretical mole ratio of sodium bicarbonate to acetic acid for the reaction? Put in the table. 5) Find the number of moles of baking soda used in each reaction (NaHCO3 = 84.0 g/mol). Write your finding on the table. 6) Determine the experimental mole ratio of sodium bicarbonate to acetic acid for each reaction (there are 8.3 x 10 moles of acetic acid in 10.0 mL of it). Fill out the table. 7) Comparing theoretical & experimental mole ratios and identify the chemical that was in excess and the limiting reactant in each test tube. Fill out the table. 8) Calculate the number of moles of CO2 formed in each test tube. Complete the table 9) Are your results agreed with your prediction? Explain.
The Correct Answer and Explanation is :
Let’s break down the lab step-by-step, using the information from the procedure. I’ll walk you through the process for each part, followed by the necessary calculations and explanations.
1) Balanced Molecular Equation
The reaction between baking soda (sodium bicarbonate, NaHCO₃) and acetic acid (C₂H₄O₂) is as follows:
[
\text{NaHCO}_3 (aq) + \text{CH}_3\text{COOH} (aq) \rightarrow \text{NaCH}_3\text{COO} (aq) + \text{H}_2\text{O} (l) + \text{CO}_2 (g)
]
This equation shows that one mole of sodium bicarbonate reacts with one mole of acetic acid to produce one mole of carbon dioxide gas.
2) The Identity of the Gas
The gas produced in this reaction is carbon dioxide (CO₂). This is evident from the molecular equation and the fact that the balloon inflates due to the production of CO₂.
3) Theoretical Mole Ratio of Sodium Bicarbonate to Acetic Acid
From the balanced equation, we can see that the mole ratio of sodium bicarbonate to acetic acid is 1:1. This means for every mole of sodium bicarbonate, one mole of acetic acid is required for complete reaction.
4) Moles of Baking Soda Used
We will calculate the number of moles of sodium bicarbonate for each test tube using the formula:
[
\text{moles of NaHCO}_3 = \frac{\text{mass of NaHCO}_3}{\text{molar mass of NaHCO}_3}
]
The molar mass of NaHCO₃ is 84.0 g/mol. Let’s calculate this for each test tube:
- Test Tube 1:
[
\frac{0.20 \, \text{g}}{84.0 \, \text{g/mol}} = 0.00238 \, \text{mol}
] - Test Tube 2:
[
\frac{0.40 \, \text{g}}{84.0 \, \text{g/mol}} = 0.00476 \, \text{mol}
] - Test Tube 3:
[
\frac{0.50 \, \text{g}}{84.0 \, \text{g/mol}} = 0.00595 \, \text{mol}
] - Test Tube 4:
[
\frac{0.70 \, \text{g}}{84.0 \, \text{g/mol}} = 0.00833 \, \text{mol}
] - Test Tube 5:
[
\frac{1.00 \, \text{g}}{84.0 \, \text{g/mol}} = 0.01190 \, \text{mol}
] - Test Tube 6:
[
\frac{1.70 \, \text{g}}{84.0 \, \text{g/mol}} = 0.02024 \, \text{mol}
]
5) Moles of Acetic Acid
The concentration of acetic acid is given as 5%, which means 5 g of acetic acid in 100 mL of vinegar. We can calculate the number of moles in 10 mL of vinegar:
[
\text{moles of acetic acid} = \frac{\text{mass of acetic acid}}{\text{molar mass of acetic acid}} = \frac{0.5 \, \text{g}}{60.0 \, \text{g/mol}} = 0.00833 \, \text{mol}
]
6) Experimental Mole Ratio
Since we know the moles of NaHCO₃ and acetic acid, we can calculate the experimental mole ratio for each test tube. This is simply the ratio of the moles of NaHCO₃ to the moles of acetic acid.
For example:
For Test Tube 1, the mole ratio is:
[
\frac{0.00238 \, \text{mol NaHCO}_3}{0.00833 \, \text{mol acetic acid}} = 0.285
]
We can calculate this for each test tube in a similar manner.
7) Limiting Reactant and Excess Reactant
To identify the limiting reactant, compare the number of moles of NaHCO₃ and acetic acid. The reactant that runs out first will be the limiting reactant. Given that the mole ratio of NaHCO₃ to acetic acid is 1:1, the reactant with fewer moles will be the limiting reactant.
For example, in Test Tube 1, the amount of NaHCO₃ (0.00238 mol) is less than the amount of acetic acid (0.00833 mol), so NaHCO₃ is the limiting reactant.
8) Moles of CO₂ Produced
Since the balanced equation tells us that 1 mole of NaHCO₃ produces 1 mole of CO₂, the moles of CO₂ produced will be equal to the moles of NaHCO₃ used in each test tube.
9) Results Analysis and Conclusion
After performing the calculations and filling out the data table, we will rank the test tubes in terms of the amount of CO₂ produced. The test tubes with less sodium bicarbonate will likely produce less CO₂ because there is less of the limiting reactant available to produce the gas. The test tube with the highest mass of baking soda should theoretically produce the most CO₂, assuming sufficient acetic acid is available. However, once the acetic acid is used up, additional sodium bicarbonate will not produce more CO₂.
This lab helps to visualize the concept of limiting reagents by showing how the reactant in lesser quantity determines the amount of product (CO₂) produced in the reaction.