Posted by Jena on Friday, October 19, 2012 at 10:08pm.
Performing the same experiment described on your lab manual, a student used a 0.358 g sample of Alka-Seltzer and the mass of CO2 was found to be 0.102 g. Based on this result, answer the following questions:
Determine the mass of NaHCO3 that produced the CO2 in the experiment.
Determine the % mass of NaHCO3 in the sample.
Determine the mass of NaHCO3 in the tablet assuming the mass of the tablet is 3.50 g.
How do i do the second and third questions?
- Chemistry - DrBob222, Friday, October 19, 2012 at 10:19pm
I answered this same question for someone earlier but I can't find it. I think I erred in that post because I don't know the reaction. What are the conditions under which the loss of CO2 occurred?
Is that a titration experiment or was the tablet heated? If heated, at what T?
- Chemistry - DrBob222, Friday, October 19, 2012 at 10:33pm
OK. I found the other post and you're the one who posted it. Therefore, disregard that post until you answer the questions about how the tablet was treated.
- Chemistry - Jena, Friday, October 19, 2012 at 10:35pm
reaction between NaHCO3 and HCl
- Chemistry - Jena, Friday, October 19, 2012 at 10:38pm
I did the equation of the % mass but its wrong
- Chemistry - DrBob222, Friday, October 19, 2012 at 10:43pm
I want to know how the tablet was treated. You make reference in the problem to "Performing the same experiment described on your lab manual, a student used a 0.358 g sample of Alka-Seltzer ......." So you know how the sample was treated but I don't. Was it titrated with something such as HCl? heat to drive off CO2. If so what temperature was used.
- Chemistry - Jena, Saturday, October 20, 2012 at 12:16am
HCl is added to NaHC03 .
- Chemistry - DrBob222, Saturday, October 20, 2012 at 12:43am
HCl + NaHCO3 => H2O + CO2 + NaCl
Convert the 0.102 g CO2 to g NaHCO3.
0.102g CO2 x (molar mass NaHCO3/molar mass CO2) = 0.102 x 84/44 = 0.195.
Is that 0.195 you posted what you obtained.
Then %NaHCO3 = (0.195/0.358)*100 = about 54.4%. Check that.
If your earlier reply meant you tried these numbers and the data base said it was wrong then I need the procedure written out for me. This answer is right for what you've posted.
- Chemistry - Jena, Saturday, October 20, 2012 at 3:53am
Yes its correct, but how about he mass of NaHCO3 in the tablet assuming the mass of the tablet is 3.50 g
- Chemistry - DrBob222, Saturday, October 20, 2012 at 3:36pm
That must be for a part of the problem you didn't post.
If the tablet is 54.4% NaHCO3 and the tablet has a mass of 3.50 g, then plug these numbers into the % formula and calculate mass NaHCO3.
(mass NaHCO3/mass tab)*100 = %
(x/3.50)*100 = 54.4
Solve for x = mass NaHCO3.
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