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Gases: Law of Combining Volumes (Worksheet)

  • Page ID
    20175
  • Q1

    Since 2 vol of acetylene require 5 volumes of oxygen, 500 c.ft will require 500 X 5/2 = 1250 c.ft of oxygen.

    For 21 volumes of oxygen we require 100 volumes of air.

    ∴ for 1250 c.ft of oxygen,

    1250 X 100 / 21 = 5952.38 c.ft of air is required

    Q4

    In a combustion chamber containing 5 L of carbon monoxide and 2.5 L of oxygen is ignited at 298 K and 1 atmosphere pressure. Assuming complete combustion and no loss of gas, what will be the volume of carbon dioxide formed at 298 K and 1 atmosphere pressure?

    Solution:

    Carbon monoxide combines with oxygen according to the equation CO + O2 → CO2.

    The balanced equation will be 2CO + O2 → 2CO2.

    Thus 2 volumes of carbon monoxide and 1 vol of oxygen combine to form 2 volumes of carbon di oxide.

    The simple ratio of 2:1:2 which will follow the Gay lussac's law of combining volumes.

    since all the gases are at the same temperature and pressure 5 L of carbon monoxide and 2.5 L of oxygen will give 5L of Carbon dioxide.

    Q5

    Q6

    What volume of propane \(C_3H_{6(g)}\) reacts with \(100 \;cm^3\) of oxygen used in the complete combustion reaction, assuming no changing pressure and temperature: (hint: construct the balanced combustion reaction).

    \[ C_3H_{6(g)} + 5O+2 \rightarrow 3CO_2 + 4H-2O\]

    Solution:


    5 vol of oxygen are used to burn = 1 vol of propane
    or 5cm3 of oxygen are used to burn = 1cm3 of propane
    1cm3 of oxygen used to burn = $\frac{1}{5}$ cm3 of propane
    100cm3 of oxygen are used to burn = $\frac{1}{5} \times 100$ cm3 of propane
    = 20cm3 of propane

    Q7

    Liquid water can be decomposed into hydrogen and oxygen under suitable conditions (e.g., electrolysis) via the balanced reaction:

    \[2H_2O_{(l)} \rightarrow 2H_{2(g)} + O_{2(g)}\]

    If an experiment generates 2500 cm3 of \(H_2\) gas, what volume of \(O_2\) is generated under the same conditions of temperature and pressure?


    Solution:
    Gay Lussac's Law is applicable to substances in the gaseous state. In the given example, Gay Lussac's Law is applicable to H2 and O2 but not to H2O.

    2H2O(l) $\rightarrow$ 2H2(g) + O2(g)


    Oxygen liberated when hydrogen liberated is 2vol = 1vol
    Oxygen liberated when hydrogen liberated is 2cm3 = 1cm3
    Oxygen liberated when hydrogen liberated is 2500cm3 = $\frac{1}{2} \times 2500$ = 1250cm3

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