This is based on \"The decomposition of Potassium chlorate\" its for my chem lab

Question

This is based on "The decomposition of Potassium chlorate" its for my chem lab and im stuck on what to do next. Help please, thank you!

Results Value Units First Decomposition RTest tube mass 19.34 g 20.31 g Test Tube KC103 mass KCIO3 mass 0.97 g Water temperature 26.5 C Water volume 225 mL 20.02 g Test tube product mass. Product mass Splint Observation Pressure of gas in system Water vapor pressure Pressure of oxygen in system Volume of oxygen in system Temperature of oxygen in system Moles of oxygen produced Moles KCIO3 used L 02/KCIO3 mole ratio Second Decompositior Test tube mass 19.97 g 20.99 Test Tube KC103 mass g KCIO3 mass 1.02 g Water temperature 26.5 C 248 mL Water volume Test tube product mass. 20.66 g Product mass Splint Observation Pressure of gas in system Water vapor pressure Pressure of oxygen in system Volume of oxygen in system Temperature of oxygen in system Moles of oxygen produced Moles KCIO3 used L 02/ KCIO3 mole ratio Standard deviation ratio Mean O2/KCIO3 mole Questions Write chemi cal equations for the t hree ossible decomposition reactions specifica for potassi um chlorate KCIO3 ib. Using the masses of KCIO3 you used in each run, calculate the mass of solid product in grams you would expect from each of the reactions above: 2a. Based on the the results m Part ID w re occurred in our test tube? Explain. fro hich action 2b. What do you conclude about the effect of the absence of the catalyst on the outcome of the thermal decomposition of KCIO3?

Explanation / Answer

a) the decomposition reactions of KClO3 are

2KClO3(s) --> 2KCl(s) + 3O2(g)

2KClO3 ---> 2KClO2 + O2 [potassium chlorite]

KClO3 ---> KClO + O2   [potassium hypochlorite]

b) we have taken 0.97 g of KClO3 initially

Moles of KClO3 taken = 0.97 / Mol wt = 0.97 / 122.55 g mol1 = 0.00791 moles

(i) for First reaction one mole of KClO3 will give one mole of KCl, therefore moles of KCl = 0.00791 moles

Mass = Moles X molecular weight = 0.0791 X 74.55 = 0.589 grams

(ii) for second reaction one mole of KClO3 will give one mole of KClO2, therefore moles of KClO2 = 0.00791 moles

Mass = Moles X molecular weight = 0.00791 X 106.55 = 0.843 grams

(iii) for third reaction one mole of KClO3 will give one mole of KClO, therefore moles of KClO = 0.00791 moles

Mass = Moles X molecular weight = 0.00791 X 90.55 = 0.716 grams

c) test tube mass = 19.34

Test tube + product mass = 20.02

Product mass = 20.02-19.34 = 0.68 grams near to mass of KClO

d) in absence of catalyst , the MnO2 , which facilitates the removal of oxygen from KClO3 hence less amount of O2 will be removed and we will obtain chloriet or hypochlorite

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