A Student thermally decomposed a 0.810 gram sample of impure potassium chlorate.

Question

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 eduiometer. Potassium chloride was the other product of the reaction. The temperature and pressure at collection time were 20.00 C and 762.10 Torr respectively. The water level in the eudiometer was 4.22 cm above the outside water level in the beaker.

What is the percent purity of the original potassium chlorate sample?

I got 40.09 % but that is not the answer I don't think.

Explanation / Answer

The balanced equation for the reaction is 2 KClO3 ------> 2 KCl + 3 O2

The vapor pressure of water at 20oC is 17.535 mm (torr). This value is adding to the observed value of oxygen gas and hence it has to be substracted.Also mm of water has to be converted to mm of mercury by dividing it with density of mercury (13.534 ).

4.22 cm = 42.2 mm

mm of mercury = 42.2 mm of water/13.534 = 3.118 mm (torr)

P = 762.10 torr - 3.118 torr - 17.535 torr = 741.447 torr

Calculate the volume of Oxygen gas collected.

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

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

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