# Calculate mass of NaHCO3 when losing CO2 in experiement

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1. Oct 8, 2015

### ptownbro

We have a homework problem that we are not sure is right. Need some help.

1. The problem statement, all variables and given/known data

The class ran an experiment where they were given 1.99 grams of NaHCO3 and 10.13 grams of HCl to create NaCl, H2O, and CO2. Before starting the experiment, what they were given weighed 12.12 grams in a beaker (which adds up to the 1.99g + 10.13g given, so that makes sense). However, after the experiment was ran, they ended up with 11.77 grams of the final product. They were told that the 0.35 grams was "lost" as CO2.

Here are the questions.
a) Calculate the mass of NaHCO3 used.
b) Calculate the % yield.

2. Relevant equations

NaHCO3 + HCl = NaCl + H2O + CO2.

3. The attempt at a solution

In our attempt to answer the questions, we first assumed the 11.77 grams only related to the NaCl + H2O part of the above equation and that the 0.35 grams relates to the CO2. Even though they say it was "lost" that just means it wasn't captured in the beaker - it's not really lost. What's left in the beaker, and what they can weigh, is just the NaCl + H2O which weights 11.77 grams.

Given that, we can use the 0.35 grams of CO2 to calculate the grams of NaHCO3 (using conversions and ratios).

a) The mass of NaHCO3 would be: We think it's 0.6682 g, but need help confirming.

0.35 g CO2 x ( 1 mole CO2 / 44 g CO2 ) x ( 84 g NaHCO3 / 1 mol NaHCO3) = 0.6682 g NaHCO3

b) The % yield would be: Not sure. Need help answering and understanding

We assumed the NaHCO3 is the limiting reactant, because at 10.13 grams of HCl would require 23.6 grams of NaHCO3 in a balanced equation (we got that by 10.13 x 1/36 x 84). Since we only have 1.99 grams of NaHCO3 that limits the HCl to 0.833 grams and the rest is excess.

However, where we got confused is if there is excess NaHCO3, where is it in the right side of the equation? Wouldn't there be a certain amount leftover there?

2. Oct 8, 2015

### Staff: Mentor

Yes, that's the amount of NaHCO3 required to produce 0.35 g of CO2.

It is not on the right side of the equation as the equation describes only the reaction- it doesn't say anything about things that didn't react.

There is a problem with the question though. How do you weight a total of 12.12 g in the beaker, it the reaction starts the moment things are mixed? You need some time to weight them, before you end the procedure, a lot of CO2 is already lost, as the reaction - especially initially - is quite vigorous.

Also, you didn't start with 10.17 g of HCl. It was 10.17 g of the HCl solution of unknown HCl concentration. Not knowing how much HCl was there you can't calculate % yield. You can calculate what fraction of the NaHCO3 reacted producing CO2, but this is not the % yield.

3. Oct 8, 2015

### ptownbro

Ah. Excellent. Makes sense. Thanks.

Yes. You're right. I was not in class room when expirement was done (this is for my daughter). So I described that detail wrong. I confirmed with her that this was done separately.

Hmm. Not sure what you mean here. Isnt the HCL is the solvent?

Also was my logic about the limiting reactant and its amount correct

4. Oct 8, 2015

### Staff: Mentor

No. Hydrochloric acid - as used in labs - is just a water solution of HCl. In "normal" conditions (pressure and temperature like the ones we typically deal with, sometimes called RTP - room temperature and pressure) HCl is a gas. Nasty gas.

As far as I an tell they were correct in terms of your understanding of what the limiting reagent is, but incorrect as they were based on incorrect assumptions about the amount of HCl. As you have calculated the amount of NaHCO3 that reacted to be 0.67 g, apparently it was in excess and it was hydrochloric acid that was a limiting reagent. Using this information it is not difficult to calculate the concentration of HCl to be 0.8 M and it is a reasonable value for a school lab experiment - concentrated enough to be still a valuable reagent for simple experiments, diluted enough to be quite safe.

5. Oct 8, 2015

### ptownbro

Ok. Thanks so much for your help. We'll try to figure it out from here. Unfortunately her teacher isn't the best at providing help and doesn't provide the final answer for homework so we don't always know if we are actually on right track with our logic.