Perform all calculations to the correct number of significant figures and show a

Question

Perform all calculations to the correct number of significant figures and show all work. Vapor pressures of water are given in the table in the experiment.

Question:1
If an Alka-Seltzer tablet contained 316 mg of aspirin, 1.927 x 1003 mg of sodium hydrogen carbonate (also known as sodium bicarbonate), and 1.0008 x 1003 mg of citric acid and nothing else, what is the percent, by mass, of sodium hydrogen carbonate in the tablet?






Question:2
Using the tablet information from Question 1, calculate the moles of CO2 you would theoretically expect from 0.273 grams of an Alka-Seltzer tablet at 720. torr and 286 K.






Question:3
You conduct an experiment where N2 is collected over water in a 2.33 L flask at 18oC. At the conclusion of the experiment, the total pressure in the flask is 412 torr. What is the pressure of N2 in the flask?






Question:4
A student reacted a sample of NaHCO3 with acid under the same conditions that will be used in lab. The total volume of gas collected plus the correction factor was 53.8 mL at a total pressure of 674 torr and 21oC. How many moles of CO2(g) were produced?

Explanation / Answer

1) mass percent of sodium bicarbonate = ( mass of sodium bicarbonate / total mass ) x 100

total mass = mass of sodium bicarbonate + mass of aspirin + mass of citric acid

total mass = 316 + 1927 + 1000.08

total mass = 3243.8

mass percent of sodium carbonate = ( 1927 / 3243.8 ) x 100

mass percent of sodium carbonate = 59.40

so the mass percent of sodium carbonate = 59.40 %

2) mass of sodium carbonate = 1927 x 273 / 3243.8

mass of sodium carbonate = 162.177 mg

moles iof sodium carbonate = mass /molar mass

moles of sofdium carbonate = 162.177 x 10-3 / 84

moles of sodium carbonate = 1.93 x 10-3

mass of citric acid = 1000.8 x 273 / 3243.8

mass of citric acid = 84.22788 mg

moles of citric acid = , 84.22788 x 10-3 / 192

moles of citric acid = 0.43868 x 10-3

the reaction is given by

C6H8O7 + 3NaHCO3 ? 3H2O + 3CO2 + Na3C6H5O7

moles of sodium bicarbonate required = 3 x moles of citric acid

moles of NaHCO3 required = 3 x0.43868 x 10-3

moles of NaHCo3 required = 1.316 x 10-3

but 1.93 x 10-3 moles of NaHCo3 is present , so NaHCO3 is in excess

moles of C02 formed = 3 x moles of citric acid

moles of C02 formed = 3 x 0.43868 x 10-3

moles of C02 formed = 1.316 x 10-3

this is at standard conditions 273 K and 760 torr , we have to find it at 286 K and 720 torr

using

P1/ n1T1 = P2 / n2 T2

n2/n1 = P2 T1 / P1 T2

n2/n1 = 720 x 273 / 760 x 286

n2/n1 = 0.904

n2 = 0.904 n1

n2 = 0.904 x 1.316 x 10-3

n2 = 1.19 x 10-3

so 1.19 x 10-3 moles of C02 is formed

4) Use

PV = nRT

n = PV / RT

n = 674 x 53.8 / 760 x 1000 x 0.0821 x 294

n = 1.976 x 10-3

so 1.976 x 10-3 moles of C02 is produced

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