A student thermally decomposed a 0.150 gram sample of impure potassium chlorate. Manganese dioxide was used as a catalyst in the reaction. T

Question

A student thermally decomposed a 0.150 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 oC and 762.10 mm Hg respectively. 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?

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Kim Chi 5 years 2021-08-28T19:07:21+00:00 1 Answers 127 views 0

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  1. thuhuong
    0
    2021-08-28T19:08:54+00:00
    Reply

    Answer:

    1. Vapor pressure of dry oxygen gas = 747.68 torr

    2. Volume at STP = 39.97 mL

    3. Number of oxygen gas molecules = 1.074 × 10²¹ molecules

    4. Percent purity of KClO3 = 97.3 %

    Explanation:

    The balanced equation for the reaction is given below :

    2 KClO3 (s) ——> 2 KCl (s) + 3 O2 (g)

    1) Since the water level in the eudiometer was below the outside water level in the beaker,

    Vapor pressure of dry oxygen gas = Total pressure + pressure due to difference in water levels – vapor pressure of water

    Vapor pressure of water at 20 °C is 17.535 mm (torr).

    Pressure due to difference in water level = 4.22 cm × 10mm/cm / 13.534 (13.534 is the density of mercury) = 3.118 mm (torr).

    Vapor pressure of dry oxygen gas = 762.10 torr + 3.118 torr – 17.535 torr

    Vapor pressure of dry oxygen gas = 747.68 torr

    2) P₁ = 747.68 torr; V₁ = 43.60 ml; T1 = 20 °C + 273.15 = 293.15 K

    P₂ = 760 torr; T₂ = 273.15 K; V₂ = ?

    Using the general gas equation = P₁V₁/T₁ = P₂V₂/T₂

    V2₂= P₁V₁T₂ / P₂T₁

    V₂ = (747.68 × 43.60 × 273.15 ) / (760 × 293.15)

    V₂ = 39.97 ml

    Volume of dry oxygen gas at STP = 39.97 mL

    3) Volume of oxygen gas at STP 39.97 mL = 0.03997 L

    Number of moles of oxygen gas in 0.03997 L = volume of gas at STP /molarvolume at STP

    Number of moles of oxygen gas = 0.03997/22.4 L

    Number of molecules of oxygen gas = 0.03997/22.4 L × 6.03 × 10²³ molecules

    Number of oxygen gas molecules = 1.074 × 10²¹ molecules

    e) Number of moles of oxygen gas = 0.03997/22.4 = 0.001784 moles

    From the equation, mole ratio of oxygen gas and potassium chlorate is 3 : 2

    Moles KClO3 = 2/3 × 0.001784 moles = 0.001189 moles

    Molar mass of KClO3 = 39 + 35.5 + 3 × 16 = 122.5 g

    Actual mass of KClO3 decomposed = 122.5 grams × 0.001189 mole = 0.146 grams

    Percent purity = (actual mass KClO3 decomposed / sample mass of impure KClO3) × 100%

    Percent purity = (0.146/0.150) × 100% = 97.3 %

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