1) What is the corrected pressure of the dry oxygen gas? 2) What is the volume i
ID: 852019 • Letter: 1
Question
1) What is the corrected pressure of the dry oxygen gas?
2) What is the volume in mL, of the dry oxygen gas at STP conditions?
3) How many molecules of oxygen were collected?
4) what is the percent purity of the original potassium chlorate sample?
A student thermally decomposed a 0.810 gram sample of impure potassium chlorate. Manganese dioxide was used as a catalyst in the reaction. The student collected 43.60 mL of oxygen gas over water in a eudiometer. Potassium chloride was the other product of the reaction. The temperature and pressure at collection time were 20.00 and 762.10 Torr respectively. The water level in the eudiometer was 4.22cm above the outside water level in the beaker. 1) What is the corrected pressure of the dry oxygen gas? 2) What is the volume in mL, of the dry oxygen gas at STP conditions? 3) How many molecules of oxygen were collected? 4) what is the percent purity of the original potassium chlorate sample?Explanation / Answer
a) Write a balanced equation for the reaction.
2 KClO3 ------> 2 KCl + 3 O2
b) What is the corrected pressure of the dry oxygen gas?
You also need to look up the water vapor pressure at 20 C in a chart as listed below. It is 17.535 mm (torr) at 20 C. Subtract this value since water vapor is adding to the observed pressure of the oxygen gas. Also, to convert mm of water to mm of mercury, divide by 13.534 (the density of mercury) to correct for the difference in water levels: (4.22 cm X 10mm/cm) / 13.534 = 3.118 mm (torr). Also subtract this value since there would have been a siphoning factor on the trapped oxygen.
P = 762.10 torr - 3.118 torr - 17.535 torr = 741.447 torr
c) What is the volume in mL of the dry oxygen gas at STP conditions?
P1 = 741.447 torr
V1 = 43.60 ml
T1 = 20 C +273.15 = 293.15 K
P2 = 760 torr
T2 = 273.15 K
V2 = ?
V2 = P1V1T2 / P2T1 = (741.447 torr)(43.60 ml)(273.15 K) / (760 torr)(293.15 K) = 39.635 ml
Answer: 38.635 ml of O2 at STP (I am concerned about significant figures, however, because only one significant figure in "20 Celsius")
d) How many molecules of oxygen gas were collected
(38.635 ml)(1.00 liter / 1000 ml)(6.022 Exp 23 molecules / 22.4 liters) = 1.0383 Exp 21 molecules
Answer: 1.0383 Exp 21 oxygen molecules
e) What is the percent purity of the original potassium chlorate sample?
Number of moles of O2 = (39.635 ml)(l.00 liter / 1000ml)(1mole / 22.414 liters / mole) = 0.001768 mole
From the balanced equation, there is a 3 : 2 mole ratio between oxygen gas and potassium chlorate. So:
Moles KClO3 = 3/2(moles O2) = 3/2(0.001768 mole KClO3) = 0.00265 moles
Molecular mass KClO3 = 39.0983 + 35.4527 + 3(15.9994) = 122.5492 grams / mole
Actual mass of KClO3 decomposed = (122.5492 grams / mole)(0.00265 mole) = 0.3247 grams
Percent purity = (actual mass KClO3 decomposed) / (sample mass of impure KClO3) X100 = (0.3247 g) / (0.810 g) X100 = 40.09% pure
Note: You may wish to adjust the significant figures in this problem because of only one significant figure in the temperature measurement.
Source:
http://genchem.rutgers.edu/vpwater.html
http://en.wikipedia.org/wiki/Mercury_%28...
http://wwwchem.csustan.edu/chem1102/mola...
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