You need to weigh your own large Pyrex test tube and know the empty mass, mass w
ID: 1032697 • Letter: Y
Question
You need to weigh your own large Pyrex test tube and know the empty mass, mass with KCIO, mixture before heating, mass with KCIO, mixture and catalyst and mass of all of the above after heating. To weigh the test tube, place it in an empty beaker to keep it upright. You may tare the weight of the beaker, b you need to know the weight of the test tube for your calculations Procedures: Weigh your clean, dry empty Pyrex test tube (1) Obtain the unknown KCI After weighing unknown sample, go to the unknown and then weigh the test tube again. (3) After thoroughly heating the mixture and no more water is being displaced, weigh the test tube mixture (4) Unlined data is obtained indirectly for your calculations o, mixture from the storeroom and weigh out about 1.4 grams in your test tube (2) storeroom to obtain the catalyst, MnO, Mix it thoroughly with the ,5782 .95 20 31.974l Data 1,9,%2 1. Weight of your empty Pyrex test tube 30.4245 grams 2. Weight of KCLO, mixture and your test tube Mass of the unknown mixture of KCLOs (2-1) 3. Weight of the unknown mixture, test tube with thoroughly mixed catalyst 31.0280 4. Weight of the unknown mixture, test tube and catalyst after heating 31.77 Mass lost by heating (also mass of gas produced) (3-4) 5. Temperature of the gas produced 25,83 oC 293.49 Kelvin 6. Volume of the gas collected 7. Barometer reading (uncorrected at 8. Corrected barometric pressure due to-mercury expansion 9. Vapor pressure of water at the temperature of the gas (from the table) 2.50 2g mL C) mm Hg mm Hg mm Hg 3.0 Vapor pressure of the oxygen collected (&.9)Explanation / Answer
1. Calculate the weight of the mixture use:
Weight of the mixture would be the difference of the masses of the mixture + tube and mass of the tube
Mass of the mixture = (mas of the mixture + tube) - mass of the tube
= 31.8990 g - 30.4295 g = 1.4695 g of unknown
2. Calculate the grams of KClO3 present in the mixture using stoichiometry of decompostion:
The decomposition reaction of KClO3 would be:
KClO3 --------------------> KClO + O2
Per the reaction, for every 1 mole of O2 produced, 1 mole of KClO3 is consumed.
So if 0.00869 moles of O2 areproduced per the calculations above, 0.00869 moles of KClO3 would be consumed.
gms of KClO3 = moles of KClO3 x molar mass of KClO3
gms of KClO3 = 0.00869 moles of KClO3 x 122.55 g/mol = 1.0650 gm KClO3
3. Calulate experimental weight % KClO3
experimental weight % KClO3= (mass determined by mole ratio / Mass of KClO3 used ) * 100%
= (1.0650 g / 1.4695 g ) * 100%
= 72.5 %
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