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Jess8219 Jess8219
wrote...
Posts: 31
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4 years ago
Hi! There was some homework questions related to an experiment that I had some trouble with, I also wanted to see if I'm correct with some of them.
So the purpose of this experiment is to determine the actual mass of sodium hydrogen carbonate in an alka-seltzer tablet and compare it with the mass given on the product label. (the following mass data from the experiment is below)
Objects used:
-Seven 250 ml Erlenmeyer Flasks
-50 ml graduated cylinder
-Dropper
-Vinegar (is dilute ethanoic acid, HC2H3O2(aq))
-7 Alka-Seltzer tablets
Here are the questions
1. Identify the dependent variable in this experiment.
2.Identify the independent variable in this experiment. 
3.Identify 2 controlled variables in this experiment. 
4. determine the mass of the sodium hydrogen carbonate that reacted in each of the 7 trials of the investigation.  In the data, you can easily determine the mass of carbon dioxide released because it is the cause of the difference in mass between the beginning of the reaction and the end.  In order to determine the mass of the sodium hydrogen carbonate that reacted, you have to use stoichiometry to compare the CO2 product to the NaHCO3 reactant. The chemical equation for the reaction of vinegar (HC2H5O(aq)  with sodium hydrogen carbonate is:                                                          NaHCO3(aq) + HC2H5O(aq)Rightwards Arrow NaC2H5O(aq) + H2O(l) + CO2(g)   
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wrote...
Staff Member
4 years ago
Is it just me or are these two columns the same thing?


- Master of Science in Biology
- Bachelor of Science
Jess8219 Author
wrote...
4 years ago
oh so sorry, my bad the first column is supposed to say BEFORE the reaction
wrote...
Staff Member
4 years ago
I figured. So from my understanding, when Alka-Seltzer reacts with acetic acid (ethanoic acid) and water as opposed to stomach acid, carbon dioxide, water, and sodium acetate are produced. Specifically, it's the sodium bicarbonate NaHCO3 that reacts with the acetic acid.

To answer the first question, the difference in mass between the two columns (after minus before) can be considered the dependent variable; essentially, that tells us how much CO2 was released. OR, you can say the mass of the flask after the reaction is the dependent variable. It's up to you.

NaHCO3 + HC2H3O2 > H2O + CO2 + NaC2H3O2

As you can see from above, the CO2 depends on the volume of ethanoic (Acetic acid).
That answers part 2.

Quote
3.Identify 2 controlled variables in this experiment. 

Not sure, wasn't there? Was time and temperature the same? That's a controlled variable. Also, we used 1 tablet every time, that's definitely a controlled variable

The document below should provide more insight~
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- Master of Science in Biology
- Bachelor of Science
Jess8219 Author
wrote...
4 years ago
I'm reviewing all of the questions and I don't think I got question 4 correct, what would be the answer to that?
Jess8219 Author
wrote...
4 years ago
please help me with this question, I'm stuck
wrote...
Educator
4 years ago
Quote
Determine the mass of the sodium hydrogen carbonate that reacted in each of the 7 trials of the investigation. 

In the data, you can easily determine the mass of carbon dioxide released because it is the cause of the difference in mass between the beginning of the reaction and the end.  In order to determine the mass of the sodium hydrogen carbonate that reacted, you have to use stoichiometry to compare the CO2 product to the NaHCO3 reactant.

The chemical equation for the reaction of vinegar HC2H5O(aq)  with sodium hydrogen carbonate is:

NaHCO3(aq) + HC2H5O(aq) <-> NaC2H5O(aq) + H2O(l) + CO2(g)

Notice that there's a 1 to 1 mole ratio of CO2 to NaHCO3, so whatever the moles of CO2 are, it's equivalent to that of NaHCO3.

I'll show you one of them, then you can do the rest:

Mass of CO2: 273.24 - 272.56 = Before - After = 0.68 g
Molar mass of CO2: 44.01 g/mol (this can be found using the periodic table)
To find the moles of CO2, multiply 0.68 by 1/44.01 = 29.9268 = 0.015 moles

This means NaHCO3 has 0.015 moles. Find the molar mass of NaHCO3, it should be 84.007 g/mol.

Multiply 0.015 moles to 84.007 g/mol to get its mass: 1.297 or approximately 1.30 grams of CO2 released.

Jess8219 Author
wrote...
4 years ago Edited: 4 years ago, Jess8219
so overall do you have to take the NaHCO3 and use the mass of co2 from each trial of the trials? so like for example the next one would be to subtract 273.49 from 272.64, and then from there you would use the molar mass of co2 and find the grams of co2 released.
wrote...
Educator
4 years ago
Yes
Jess8219 Author
wrote...
4 years ago
okay so the answer to the next one would be 1.60 grams of CO2 released?
wrote...
Educator
4 years ago
That's right, but I get 1.62235. Rounded to 1.62. Where'd you get 1.60?
Jess8219 Author
wrote...
4 years ago
my calculator says 1.596133
wrote...
Educator
4 years ago
Maybe when you divided the difference in the before/after calculation, then divided by 44.01, you didn't carry all the decimals. Make sure you don't round prematurely in the calculations, only round @ the end
Jess8219 Author
wrote...
4 years ago
I didn't round anything, I only rounded at the end but it wouldn't be considered wrong?
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