Chem help: how to calculate the experimental and theoretical volume of a dry gas at STP?

In two trials 2KClO3 -> 2KCl + 3O2

with MnO4 a catalyst was heated while joined with a pneumatic trough that had 4 jars in the first and second trial filled with water. The point is to produce gas in the jars.

Trial 1:

Weight of KClO3:1.754g

Temp of water in Trough: 38.8 degrees Celsius.

weight of empty jar: 144.001g

ml of water in jar: 159.5ml

weight of contents after heating: 2.479g

3 jars empty + 1 jar filled with oxygen and ml of water left in 4th jar : 143.9 ml

Trial 2:

Weight of KClO3: 2.06g

Temp of water in Trough: 24.5 degrees Celsius.

weight of empty jar: 151.329g

ml of water in jar: 159.5ml

weight of contents after heating: 2.143g

3 jars empty + 1 jar filled with oxygen, volume of water in jar::155.9ml

atm pressure: 100.5kpa

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  • 7 years ago
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    Trial 2:

    so I think what you are telling me is that

    3 jars empty: produced (3) (159.5 ml ea) = 478.5 ml of gas measured in 3 jars

    Plus whatever was in the 4th Jar

    was it " filled with oxygen," or did it have "volume of water in jar::155.9ml"

    if it had "volume of water in jar::155.9ml", then

    (159.5 ml - 155.9 ml H2O still there in the 4th jar = another 3.6 ml gas measured

    for a total of

    482.1 ml,

    so make adjustments if the fourth jar had 155.9 ml of O2, rather than 3.6 ml of O2 gas

    ================

    from: http://msduncanchem.com/Reference_Tables/water_vap...

    the water vapor pressure at 24.5 C is 3.10 kPa

    Dalton's Law tells us that the

    total Pressure = Pressure of dry O2 + pressure of the water vapor

    so, the dry pressure would be

    100.5kPa - 3.1 kPa = 97.4 kPa

    to calculate the experimental and theoretical volume of a dry gas at STP

    P1 = 97.4 kPa

    V1 = 482.1 ml

    T1 = 24.5 C + 273 = 297.5 Kelvin

    P2 = 101.3 kPa (STP)

    V2 = ?

    T2 = 273.1 Kelvin

    P1V1T2 = P2V2T1

    V2 = (P1) (V1) (T2) / (P2) (T1)

    V2 = (97.4) (482.1) (273.1) / (101.3) (297.5)

    your answer is

    V2 = 425.5 ml of dry O2

    ==================================

    ==================================

    trial 1:

    (the temp of the water looks a bit hot)

    Temp of water in Trough: 38.8 degrees Celsius.

    ml of water in jar: 159.5ml

    3 jars empty + 1 jar filled with oxygen and ml of water left in 4th jar : 143.9 ml

    ==========

    so I think what you are telling me is that

    3 jars empty: produced (3) (159.5 ml ea) = 478.5 ml of gas measured in 3 jars

    Plus whatever was in the 4th Jar

    was it " filled with oxygen," or did it have "volume of water in jar::143.9ml"

    if it had "volume of water in jar::143.9ml", then

    (159.5 ml - 143.9 ml H2O still there in the 4th jar = another 15.6 ml gas measured

    for a total of

    494.1 ml,

    ================

    extrapolating between 38 C & 39C

    from: http://www.phs.d211.org/science/smithcw/AP%20Chemi...

    he water vapor pressure at 38.8 C is 6.9 kPa

    Dalton's Law tells us that the

    total Pressure = Pressure of dry O2 + pressure of the water vapor

    so, the dry pressure would be

    100.5kPa - 6.9 kPa = 93.6kPa

    to calculate the experimental and theoretical volume of a dry gas at STP

    P1 = 93.6 kPa

    V1 = 494.1 ml

    T1 = 38.8 C + 273 = 311.8 Kelvin

    P2 = 101.3 kPa (STP)

    V2 = ?

    T2 = 273.1 Kelvin

    P1V1T2 = P2V2T1

    V2 = (P1) (V1) (T2) / (P2) (T1)

    V2 = (93.6) (494.1) (273.1) / (101.3) (311.8)

    your answer is

    V2 = 399.9 ml of dry O2

    email me with any modifications

    p.s. if the 1.754 g of KClO3 was pure KClO3

    then we can calculate:

    the moles of KClO3, (using its molar mass)

    the moles of O2 expected, (at a ratio of 3 mol O2 to 2 mol KClO3)

    & the volume of dry O2 expected at STP (at a ratio of 22.4 Litres per mol)

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