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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