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3. Potassium perchlorate, KCIO, decomposes on heating to form potassium chloride

ID: 589011 • Letter: 3

Question

3. Potassium perchlorate, KCIO, decomposes on heating to form potassium chloride and elemental oxygen. a Write a balanced molecular equation for the thermal decomposition of Using a setup like this experiment, the following data were collected mass of sample before heating mass of residue after heating volume of water displaced atmospheric pressure water temperature 1.2460 g 0.9265 g 258 ml. 748.9 torr 26.0 C b. What is the percent mass KCIO, in the sample being heated? What volume would this sample of oxygen occupy if collected at standard temperature and pressure? c. d. What is the molar volume at STP of O, according to these data?

Explanation / Answer

a) heat

KClO4(s) ----------> KCl(s) + 2O2(g)

b)

Mass of sample before heating = 1.2460g

Mass of sample after heating = 0.9265g

loss on heat = 0.3195g

No of mole of O2 gas = 0.009984

No of mole of KClO4 = (1/2)× 0.009984 = 0.004992

Molar mass of KClO4 = 138.55g/mol

Mass of KClO4 = 0.004992mol × 138.55g/mol = 0.6916g

Mass percent of KClO4 = (0.6916g/1.246g)×100 = 55.51?

c) No of mole of Oxygen,n = 0.009984

Tempersture ,T = 26? = 299.15K

pressure ,P = 748.9torr = 0.988atm

Volume , V = nRT/P

= 0.009984× 0.082057(L atm /mol K)× 299.15K/0.988atm

= 0.2481L

d)

Standard temperature = 273.15K

Standard pressure = 1arm

V = nRT/P

= 0.009984× 0.082057( L atm / mol K ) × 273.15K /1atm

= 0.2238L

molar volume of O2 = 0.2238L/0.00998ol = 22.4L/mol

  

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