A student thermally decomposes a 0.150 g sample of impure potassium chlorate. Ma

Question

A student thermally decomposes a 0.150 g 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 C and 762.10 mmHg. The water level in the eudiometer was 4.22 cm below 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

For the experiment described above,

1) water vapor pressure at 20 oC = 17.535 torr [literature]

correct for the difference in water levels: (4.22 cm X 10mm/cm) / 13.534 = 3.118 mm (torr)

Corrected pressure of dry O2 gas = 762.10 - 17.535 - (4.22 x 10/13.534) = 741.447 torr

2) Using,

P1V1/T1 = P2V2/T2

with,

P1 = 741.447 torr

V1 = 43.60 ml

T1 = 20 oC + 273.15 = 293.15 K

P2 = 760 torr

T2 = 273.15 K

V2 = ?

So Volume of dry O2 gas at STP,

V2 = P1V1T2 / P2T1

      = 741.447 torr x 43.60 ml x 273.15 K/760 torr x 293.15 K

     = 39.633 ml

3) molecules of O2 gas collected,

= 39.633 ml x 6.023 x 10^23/22.4 L x 1000 ml

= 1.065 x 10^24 molecules

4) moles of O2 collected = 39.633 ml/22.414 x 1000ml = 0.00177 mol

moles of KClO3 = 0.00177 mol x 2/3 = 0.0012 mol

mass of KClO3 decomposed = 122.5492 g/mol x 0.0012 mol = 0.144 g

So,

Percent purity of original KClO3 sample = 0.144 g x 100/0.150 g = 96.0 %

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