experiment 5 Revised data sheet: Analysis of an unknown KCLO/ KCI mixture Unknow
ID: 1032597 • Letter: E
Question
experiment 5 Revised data sheet: Analysis of an unknown KCLO/ KCI mixture Unknown Number 22 You need to we igh 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, but 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 KCIO, mixture from the storeroom and weigh out about 1.4 grams in your test tube (2) After weighing unknown sample, go to the storeroom to obtain the catalyst, MnOj. Mix it thoroughly with 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 Data Weight of your empty Pyrex test tube Weight of KCLO, mixture and your test tube Mass of the unknown mixture of KCLO 30.0248grams 32.0l 1. 2. 1.442 g (2-11 3. Weight of the unknown mixture, test tube with thoroughly mixed catalyst 22.0al 4. Weight of the unknown mixture, test tube and catalyst after heating 31.1039 0.1832 Mass lost by heating (also mass of sas produced) (3-4) 5. Temperature of the gas produced 21./ 2ggu Kelvin 6. Volume of the gas collected ? Barometer reading (uncorrected at 16oC?7b torr 8. Corrected barometric pressure due to mercury expansion 9. Vapor pressure of water at the temperature of the gas (from the table) m mL 50m hn t mm Hg 16%,1 Hol. 77mm Hg mm Hg Vapor pressure of the oxygen collected (8-9Explanation / Answer
1) gas produced upon heating=32.0961-31.9079=.1882gms
2)partial pressure of gas produced =761.77-25.2=736.57tors
3) PV=nRT
n=PV/RT=(736.57*.164)/(62.36367*299.1)=0.00647 moles
No of moles of gas obtained is 0.00647 moles
4) grams/moles= .1882/.00647=28.76gms/mole
5)percentage of error = .1882-.1589/.1882
=.15567*100
=15.56%
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