Molar volume is the volume that one mole of gas occupies when temperature and pressure are kept constant. The molar volume of a gas can be determined through evaluating how much gas is given off when the number of moles of the substance is known. To find the volume of gas that will be used to calculate the molar volume, the process of water displacement can be used.
Cesa, J. (2002). ChemTopic labs: Experiments and demonstrations in chemistry (vol. 9). Batavia, Il: Flinn Scientific. Calculations
(Weight of Mg ribbon used for conversion) (____¬.50 g¬¬¬¬____) = .038 g/cm2 (Width of ribbon)(length of conversion Mg ribbon) (.3 cm x 44.15 cm)
(Conversion factor)(Length of Mg ribbon)(width of Mg ribbon) = mass of Mg ribbon .038 g/cm2 (.9 cm x .3cm) = .0103 g
Volume of H2 gas 11.5 mL Amount to be subtracted or removed to correct the meniscus- .2 mL Corrected volume of H2 gas 11.3 mL
Corrected volume of H2 gas converted to liters 11.3 mL (1 x 10-3 L) =.0113 L (1 mL)
Temperature of water bath in K 22.1°C + 273.15K = 295.3K
Barometric Pressure minus water vapor pressure equals pressure of H2 gas 744.72 mmHg – 19.8 mmHg = 724.9 mmHg
P1V1T2 = V2 (724.9 mmHg)(.0113 L)(273.15 K) = 9.97 x 10-3 L P2T1 (760 mmHg)(295.3 K) Volume of H2 (g) at STP
Volume of H2 gas (9.97 x 10-3 L) = 23.5 L/mol Theoretical amount of moles (4.24 x 10 -4 mol) Molar Volume
Mass of Mg ribbon times molar mass equals moles of Mg .0103 g Mg ( 1 mol ) = 4.24 x 10-4 mol Mg (24.3050 g)
Percent Yield (23.5 L/mol) x 100 = 105% (22.42 L/mol) Percent Yield
Data Table 1 Length of Mg Ribbon.9 cm Mass of Mg.0103 g Evidence of Chemical ReactionGas bubbles came up off of the iron cage containing the Mg ribbon Volume of H2 Gas 11.5 mL Corrected Volume of H2 Gas11.3 mL Temperature of Water Bath Before Reaction22.1° C Temperature of Water Bath After Reaction22.0° C Barometric Pressure744.72 mmHg
Water displacement can be used to determine the amount of a gas that a reaction exudes. That volume can then be used to calculate the molar volume of the gas after the measured volume is corrected for differences in temperature and pressure. When a metal, acid, and water are placed into a graduated cylinder, that graduated cylinder can then be inverted into a water bath. As a reaction occurs, the gas that is produced will rise to the new “top” of the graduated cylinder. This will push some of the water out of the graduated cylinder and into the water bath. The volume of gas can be determined after the reaction has run to completion by reading the amount of space the gas has taken up and subtracting .2 mL due to the inverted meniscus.
Using a copper wire, a “cage” was made around a .9 cm long piece of magnesium ribbon, which was then placed into a rubber stopper. After placing 5.0 mL of 2 M hydrochloric acid into a 25 mL graduated cylinder, distilled water was layered overtop of the acid until the water was almost brimming of the edge. The rubber stopper was put into the graduated cylinder firmly, and then quickly inversed into the water bath. The formation of gas signified that a reaction was occurring. The gas was able to be collected at the top of the graduated cylinder when it was inversed due to the pressure pushing the water out of the graduated cylinder. The results were recorded before the reaction was finished due to a time constraint.
The volume of hydrogen gas was 11.5 mL, and the corrected volume was 11.3 mL because of the inversed meniscus, and the temperature of the water bath after the reaction was 22.1 °C. Using this information, the theoretical amount of moles of H2 gas that was to be produced was found to be 4.24 x 10 -4 moles, which was calculated by converting our mass of Mg ribbon into moles of H2. Using the combined gas law we calculated the volume of H2 gas at STP. This then allowed us to find the molar volume of our lab by dividing the volume of H2 gas produced at STP by the theoretical amount of moles. Our molar volume was 23.5 L/ mole. We found our percent yield to be 105%, and this was calculated by dividing our lab’s molar volume by the theoretical molar volume. Since our percent yield cannot actually be 105%, one or more errors could have occurred to cause this issue. One error that could have occurred was wrapping the copper wire too tightly around the magnesium. This would cause the reaction to take much longer than if we had wrapped it more loosely. Due to time, we weren’t able to let the reaction completely finish. Although, we determined that the amount of gas that was left to be given off was much too small of an amount to make much of a difference. Another error was how our total barometric pressure was an average between the pressure reading in the hallway and the pressure reading in the room. Using an average would have caused a difference in our calculations because since the barometric pressure was not exact, any calculations involving this average would not be completely correct. Another error could have been if we missed a spot of oxidation on our magnesium ribbon, which then could have caused a new substance to be introduced to the reaction. This error could have caused the molar volume of hydrogen to be lower than what was to be expected because part of the Mg would have already reacted. Even if we did clean off all of the visible oxidation, this metal would have started oxidizing again immediately. One last error was if we had allowed air to get into the graduated cylinder. This could have caused a bubble to form, which would have made our measured volume too high.
1. Vapor pressure of water at 22.0°C = 19.8 mmHg
Mg (s) + HCl (aq) → H2 (g)
22.0°C + 273.2 K = 295.2 K
Ptotal = P(g) + P(H2) 746 mmHg = P(g) + 19.8 mmHg 726 mmHg = P(g)
2. P1V1T2= P2V2T1
22.0°C + 273.2 K = 295.2 K
31.0 mL (1x 10 -3 L) = .0310 L ( 1 mL )
(726 mmHg)(.031 L)(273.15 K) = .0274 L (760 mmHg )(295.2 K)
3. Mg (s) + 2HCl (aq) → MgCl2 (aq) + H2 (g)
Mg = 24.3050 g/mol
0.028 g Mg ( 1 mol ) = .0012 mol Mg (24.3050 g)
4. Corrected volume of H2 = .0274 L = 22.8 L/mol Theoretical # of moles of H2 .0012 mol
1..0103 g Mg (1 mol Mg) (1 mol H2) = 4.24 x 10 -4 mol H2 (24.3050 g) (1 mol Mg)
The theoretical number of moles of hydrogen gas produced in Trial 1 was 4.24 x 10 -4 moles.
2. 744.72 mmHg ( 1 atm ) = .97989 atm (760 mmHg)
19.8 mmHg ( 1 atm ) = .0261 atm (760 mmHg)
Ptotal = P(H20) + P(H2) .97989 atm = (.0261 atm) + P(H2) P(H2) = .9538 atm
The partial pressure of hydrogen gas that was produced was .9538 atm. 3. P1V1T2 = V2 (724.9 mmHg)(.0113 L)(273.15 K) = 9.97 x 10-3 L P2T1 (760 mmHg)(295.3 K)
9.97 x 10-3 L ( 1 mL ) = 9.97 mL (1 x 10-3 L)
The hydrogen gas would occupy 9.97 x 10-3 L or 9.97 mL
4. 9.97 mL H2 gas ( 1 x 10 -3) = 9.97 x 10-3 L ( 1 mL )
Molar Volume = (Volume of H2) (9.97 x 10-3 L H2 ) = 23.5 L / mol (Theoretical # of moles H2) (4.24 x 10-4 mol H2)
The molar volume is 23.5 L/ mol.
5. Percent error = |Experimental value – Literature value| x 100 Literature value
Percent error = |23.5 L – 22.42 L| x 100 = 4.82% 22.42 L
The percent error was 4.82%
6. 1 mol of H2 (g) (2.02 g ) (1 mol) = .0860 g/L Molar volume (1 mol ) (23.5 L )
The experimental value for the hydrogen gas was 0.860 g/L while the literature value was .0899 g/L.
7. A bubble of air in the graduated cylinder would have caused the measured volume of hydrogen gas to be too high. This would have happened because of the appearance of more hydrogen gas when the volume was read initially.
8. The error of oxidation would have caused the measured volume to be lower than it should have been due to the introduction of an extra substance (the oxidation) being added to the reaction because part of the Mg would have already reacted with the oxidation.
9. Buret mL converted to L 50. mL (1 x 10 -3L ) = .050 L ( 1 mL )
Temperature of water bath from °C to K 22.1°C + 273.2K = 295.3K
n= PV RT
n= (744.72 mmHg) (.050L) = 2.0 x 10 -3 mol (62.4 L mmHg/mol K)( 295.3K)
Converting mol of Mg to mass 2.0 x 10 -3 mol Mg ( 24.3050g) = .049 g Mg ( 1 mol )
Converting mass of Mg to length .049 g Mg (44.15 cm) = 4.3 cm ( .50 g )
The maximum length of Mg ribbon that should be used is 4.3 cm.
Cybersecurity and Infrastructure Security Agency (CISA)
Cybersecurity and Infrastructure Security Agency (CISA).
There are three graded components to the project. Strategic Plan and must use that sample plan EXACTLY as a template when submitting. Chapter 1: Introduction must contain the following items:
Cover Page Working Table of Contents Body of Chapter (5 pages of narrative) comprising: the background the statement of the problem operational definitions the significance of the study Chapter Summary References Page The body of this chapter requires five full (5) pages of narrative, which does not include photographs, bulleted lists, or other inserts. Chapter 2: The Literature Review must contain the following items: Cover Page Working Table of Contents Updated Chapter 1, revised using instructor comments Body of Chapter 2 (5 pages of narrative) comprising the pertinent studies related to the selected organization and strategic plan. References Page The body of this chapter requires five full (5) pages of narrative, which does not include photographs, bulleted lists, or other inserts. The student must choose at least six (6) pertinent studies for inclusion in the Literature Review. These studies should be related to the organization selected and cover the need to accomplish a strategic plan for this organization. Chapter 3: The Strategic Plan and Chapter 4: Conclusions and Recommendations must contain the following items: Cover Page Table of Contents Updated Chapter 1, revised using instructor comments Updated Chapter 2, revised using instructor comments Body of Chapter 3 (8.5 pages of narrative) comprising: A chapter introduction a restatement of the problem a mission and vision statement the strategic plan description a chapter summary Body of Chapter 4 (1.5 full pages of narrative) comprising: Conclusion Recommendations References Page
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