Mass of ignition tube (PDF)

EXPERIMENT 7EXPERIMENT 7
NAME:NAME:
PARTNER:PARTNER:
DATE:DATE:
REPORT SHEETREPORT SHEET
DATA TABLEDATA TABLE Your DataYour Data Trial DataTrial Data
Mass of ignition tube & contentsMass of ignition tube & contents
(before heating)(before heating) gg 48.3648.36 gg
Mass of ignition tube & contentsMass of ignition tube & contents
(after heating)(after heating) gg 47.5347.53 gg
Temperature of oxygenTemperature of oxygen
(taken in flask)(taken in flask) ooCC 23.623.6 ooCC
Volume of Oxygen collectedVolume of Oxygen collected
(volume of water displaced)(volume of water displaced) mLmL 595.8595.8 mLmL
Barometric readingBarometric reading
(directly from barometer)(directly from barometer) in.in. 30.4330.43 in.in.
Barometric PressureBarometric Pressure
(in torr)(in torr) torrtorr 772.9772.9 torrtorr
Vapor pressure water at _________Vapor pressure water at _________ooCC
(Table 6, Appendix)(Table 6, Appendix) torrtorr 21.821.8 torrtorr

CALCULATIONSCALCULATIONS Your DataYour Data Trial DataTrial Data
Mass of OxygenMass of Oxygen gg 0.83 g0.83 g
Temperature of OxygenTemperature of Oxygen (in Kelvin)(in Kelvin) KK 296.8 K296.8 K
Pressure of OxygenPressure of Oxygen (in torr)(in torr) torrtorr 772.9772.9 torrtorr
Volume of OxygenVolume of Oxygen (at STP)(at STP) LL 0.5424 L0.5424 L
Molar Volume of OxygenMolar Volume of Oxygen (from your data)(from your data) L/molL/mol 20.91 L/mol20.91 L/mol
Percent ErrorPercent Error %% 0.067 %0.067 %
Calculation
(i)(i) Mass of oxygen= Mass of ignition tube & contentsMass of ignition tube & contents (before heating)-(before heating)- Mass of ignitionMass of ignition
tube & contentstube & contents (after heating)=(after heating)= 0.83g0.83g
(ii)(ii) Temperature of oxygen (in kelvin) =273.2k+ 23.6= 296.8KTemperature of oxygen (in kelvin) =273.2k+ 23.6= 296.8K
(iii)(iii) Pressure of oxygen (in torr) =772.9 torrPressure of oxygen (in torr) =772.9 torr
(iv)(iv) Volume of oxygen (at STP)= TVolume of oxygen (at STP)= The pressure of the O2 alone is 772.9 Torr – 21.8 Torr
= 751.1 Torr
Volume of oxygen collected=Volume of oxygen collected= 595.8595.8 mLmL
(P1V1/T1)=(P2V2/T2)= (751.1*595.8/296.8)=(760*V2/273.2k)= 542.37ml(P1V1/T1)=(P2V2/T2)= (751.1*595.8/296.8)=(760*V2/273.2k)= 542.37ml
(v) The volume of O2 collected at STP is 542.37 mL or 0.5424 L

(vi) Number of moles of O2:
0.83 g*(1mole of 02/32g of O2) = 0.02594 mole of oxygen gas
Molar volume= 0.5424L02/0.02594= 20.91L/mol of oxygen gas
(vii) Percent error= (theoretical-experimental value/theoretical value)
= (22.4L/mol-20.91L/mol/22.4L/mol)= 0.067%= (22.4L/mol-20.91L/mol/22.4L/mol)= 0.067%

1. Why must the test tube be dry before adding the KClO3/MnO2 mixture in step 1?
Potassium chlorate is a strong oxidizing agent (Conkling and Mocella, 2010). Therefore, water
drops could react with potassium chlorate (KClO3). Secondly, the wet test tube will increase the
weight of the test tube, thus giving inaccurate mass.
2. What is the function of the MnO2?
To fasten the disintegration of potassium chlorate, MnO2 is needed which acts as a catalyst.
Catalyst is a material that causes an upsurge in the chemical reaction speed without being used
up in the reaction (Lewis & Hawley, 2016). Heating KClO3 in the absence of a catalyst changes
it into potassium perchlorate (Chandler, 2014).
3. Why do you elevate the beaker until the water level in the beaker and flask is the same?
Both in step 5 and step 10?
As the O2 is collected above water, water gases will also be existent in the vapor. The trial is
designed so that the overall pressure of the oxygen and water gas will be equivalent to the
atmospheric pressure (Lewis & Hawley, 2016). The purpose of raising beaker is to ensure
equalization process occur; pressure acting in the beaker should be equal to the pressure acting
on the water in the flask.
4. Why don't you dry the beaker in step 10?
Because the capacity of the water in glass is equivalent to the size of the gas produced, drying
the beaker, it will mean the volume of the oxygen obtained will be low.
EXPERIMENT 7EXPERIMENT 7
MOLE VOLUME OF A GASMOLE VOLUME OF A GAS

5. Write the balanced equation for this reaction.
2 KClO3 → 2KCl + 3O2 (in presence of MnO2)
6. Why does the water volume in the beaker equal the volume of oxygen produced?
The trial of KClO3 is positioned in the test tube and flask is occupied with water. Certain amount
of water is moved by oxygen and is pressed into the tumbler. The volume of water in the beaker
will be identical to the volume of O2 in the flask. When the test tube is heated, the solid will
thaw, oxygen will be liberated, and water from the flask will be moved into the tumbler.
Therefore, the volume of water displaced is equal to the volume of the oxygen (Rai, Dinakar,
Kurian and Bindoo, 2014). Care should be taken at first by brushing over the test tub tube.
7. What happens to the volume of water in the beaker as the apparatus cools in step 12?
The bulk of water in the glass will remain the same, since there is no heating of test tube
containing potassium chromate to produce oxygen gas.
8. What would happen if you forgot to remove the clamp during heating?
If the clamp is not unfastened, the generation of gas during heating could cause an outburst, even
though it is more probable that a plug would be enforced to slacken. Similarly, one should make
sure that the lengthier glass rod is not touching the Erlenmeyer flask bottom. Therefore, if the
clamp is not opened, it will create a closed system and an outburst will occur.