Alka Seltzer Stoichiometry Lab Answer Key

The Alka Seltzer Stoichiometry Lab Answer Key

You know this lab. Every intro chemistry class does it at some point. You drop an Alka Seltzer tablet into water, collect the CO gas, and try to prove you understand mole ratios. The answer key most teachers are working from has a very specific set of expected numbers, and honestly, your students are going to miss them. A lot. Here is what that answer key actually looks like and how to use it without losing your mind.

Where to find the Alka Seltzer Stoichiometry Lab Answer Key

The standard version this answer key refers to assumes you are using a 325 mg aspirin/acid-buffer Alka Seltzer tablet with roughly 1916 mg total mass, and the active ingredients are sodium bicarbonate (NaHCO), citric acid (CHO), and aspirin. The stoichiometry portion focuses on the reaction between NaHCO and citric acid in aqueous solution: 3 NaHCO + CHO NaCHO + 3 HO + 3 CO The molar masses are NaHCO at 84.01 g/mol, citric acid at 192.12 g/mol, and CO at 44.01 g/mol. Most keys expect students to calculate the theoretical yield of CO based on whichever reactant is limiting. In a typical tablet, NaHCO is the limiting reagent by a wide margin, which is the whole point of the exercise.

How the lab actually runs

You have students measure mass of the tablet, drop it into an Erlenmeyer flask with a balloon or gas collection setup, capture the CO, and then either measure the volume of gas or the mass loss from the system. The answer key walks through converting that gas volume (usually at room temperature and pressure) back to moles using the ideal gas law, then working backwards to find how much NaHCO reacted. Students who treat STP as 273 K and 1 atm get consistently wrong answers. The trick is using the actual room conditions in the lab. I had a whole section one year where every group got 112 percent yield because they used STP in a room that was 23°C and 752 mmHg. Once I made them plug in their actual T and P, the numbers dropped to something around 94 percent, which is actually reasonable for this kind of setup.

Common pitfalls the answer key assumes you already know about

CO is soluble in water. If students are bubbling gas through water to collect it by displacement, they are losing CO to dissolution. The answer key usually accounts for this with a small correction factor, but most teachers skip it. I just tell the kids to use the mass-loss method instead. You weigh the flask before and after the reaction, and the difference is the mass of CO that escaped. It cuts the error down to about 5 percent instead of 20. Another thing the answer key does not always flag clearly: the tablet coating. Alka Seltzer tablets have a thin film coating that adds mass but does not contribute to the reaction. If students are calculating the percent by mass of NaHCO in the tablet, they need to subtract the coating weight first. That is usually around 5 to 8 percent of the total tablet mass depending on the product line. I measured this once by dissolving a tablet, filtering the residue, drying it, and weighing it. The unreacted gunk came out to about 7.2 percent. Your answer key might say 5 percent. They are both defensible if you explain the methodology.

Sample answer key walkthrough

Let me run through one of the standard problems the way the key expects it. Say a student collects 215 mL of CO at 22.5°C and 755 mmHg. First, convert to SI-adjacent units: P = 755 mmHg × (1 atm / 760 mmHg) = 0.9934 atm V = 0.215 L

T = 22.5 + 273.15 = 295.65 K n = PV / RT = (0.9934 × 0.215) / (0.08206 × 295.65) = 0.00885 mol CO From the balanced equation, 3 mol CO comes from 3 mol NaHCO, so the mole ratio is 1:1. That means 0.00885 mol NaHCO reacted.

Mass of NaHCO = 0.00885 mol × 84.01 g/mol = 0.743 g or 743 mg If the tablet was 1916 mg total, the percent NaHCO by mass would be about 38.8 percent. Most commercial Alka Seltzer tablets list NaHCO content in the 30 to 40 percent range, so this checks out. The answer key will then ask for percent yield. If the theoretical CO from the labeled NaHCO content is 0.0102 mol but the student measured 0.00885 mol, the percent yield is 86.8 percent. That is an acceptable result for this lab.

What the answer key does not cover

Some versions of this lab use the effervescent tablets that contain sodium carbonate instead of sodium bicarbonate. The stoichiometry changes completely. The reaction becomes: NaCO + CHO 2 Na + CHO³ + HO + CO If your answer key is built for NaHCO and a student group is using NaCO tablets, every calculation will be off by a factor related to the different molar ratios and molar masses. I ran into this last spring when a supply chain issue switched our order to a different brand without updating the handout. It took an hour of panicked recalculations before I realized what happened. Just verify the tablet composition before you start.

The other gap is vapor pressure of water. If you are collecting gas over water, the total pressure inside the collection vessel is the sum of the CO partial pressure and the water vapor pressure. At 22.5°C, the vapor pressure of water is about 20.1 mmHg. So the actual CO pressure is 755 minus 20.1, which is 734.9 mmHg. Using the uncorrected pressure inflates your mole calculation by roughly 2.7 percent. The answer key sometimes includes this correction, sometimes does not. Check with your department's standard.

A practical tip for grading

Don't penalize students for using slightly different molar masses. Some textbooks round NaHCO to 84.0, others keep it at 84.01. The difference shows up in the third significant figure and can shift a final percent yield by a point or two. As long as their method is correct and they show their work, both answers deserve full credit. The answer key I have seen that grades most fairly allows a tolerance band of ±3 percent on the final yield number, which accounts for normal experimental variation without letting sloppy work slide.