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    How do you calculate the volume of oxygen required for the complete combustion of 0.25 dm^3 of methane at STP?

    The volume of oxygen required is "0.50 dm"^3. > For this problem, we can use Gay-Lussac's Law of Combining Volumes: If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers. The balanced equation for the combustion is color(white)(l)"CH"_4 + "2O"_2 → "CO"_2 + "2H"_2"O" "1 dm"^3color(white)(ll)"2 dm"^3 According to Gay-Lussac, "1 dm"^3color(white)(l) "of CH"_4 requires "2 dm"^3color(white)(l) "of O"_2. ∴ "Volume of O"_2 = 0.25 color(red)(cancel(color(black)("dm"^3 color(white)(l)"CH"_4))) × ("2 dm"^3color(white)(l) "O"_2)/(1 color(red)(cancel(color(black)("dm"^3color(white)(l) "CH"_4)))) = "0.50 dm"^3color(white)(l)"O"_2

    How do you calculate the volume of oxygen required for the complete combustion of 0.25

    d m 3 of methane at STP?

    Chemistry Gases Molar Volume of a Gas

    1 Answer

    Ernest Z. Sep 10, 2016

    The volume of oxygen required is

    0.50 dm 3 .

    Explanation:

    For this problem, we can use Gay-Lussac's Law of Combining Volumes:

    If pressure and temperature are constant, the ratio between the volumes of the reactant gases and the products can be expressed in simple whole numbers.

    The balanced equation for the combustion is

    l CH 4 + 2O 2 → CO 2 + 2H 2 O 1 dm 3 l l 2 dm 3

    According to Gay-Lussac,

    1 dm 3 l of CH 4 requires 2 dm 3 l of O 2 . ∴ Volume of O 2 = 0.25 dm 3 l CH 4 × 2 dm 3 l O 2 1 dm 3 l CH 4 = 0.50 dm 3 l O 2 Answer link

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    स्रोत : socratic.org

    Calculate the volume of oxygen required for the complete combustion of 0.25 dm^3 of methane at STP.

    Click here👆to get an answer to your question ✍️ Calculate the volume of oxygen required for the complete combustion of 0.25 dm^3 of methane at STP.

    Calculate the volume of oxygen required for the complete combustion of 0.25dm

    Question 3 of methane at STP. Medium Open in App

    Updated on : 2022-09-05

    Solution Verified by Toppr

    Given volume of methane =0.25dm

    3 =0.25L

    Combustion of methane (CH

    4 ​ )- (methane) CH 4 ​ ​ +2O 2 ​ ⟶CO 2 ​ +2H 2 ​ O

    As we know that volume of 1 mole of gas is 22.4 L.

    Amount of oxygen required for the combustion of 22.4 L of methane =(2×22.4)L=44.8L

    ∴ Amount of oxygen required for the combustion of 0.25 L of methane =

    22.4 44.8 ​ ×0.25=0.5L

    Hence, 0.5 L of oxygen required for the complete combustion of 0.25dm

    3 of methane.

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    Calculate the volume of oxygen required complete combustion ?

    Calculate the volume of oxygen required complete combustion of 0.25 mole of methane STP (Ans. 11.2 dm3) - 188538

    Scholr

    Calculate the volume of oxygen required complete combustion of 0.25 mole of methane STP (Ans. 11.2 dm3)

    1 Answers

    1 Answers Praveen Batra

    Grade 9

    The reaction for combustion of Methane proceeds as follows:

    CH4 + 2O2 ----------> CO2 + 2H2O

    Thus for 1 mole methane we need 2 mole of oxygen for complete combustion.

    So, for 0.25 mole methane the amount of oxygen in mole would be : 0.25 x 2 = 0.5 mol of oxygen

    As we know at STP conditions 1 mole of gas occupies 22.4 L

    Hence 0.5 moles of oxygen will occupy 22.4 L x  0.5 = 11.2 L.

    11.2 L of oxygen is required for complete combustion of 0.25 mole methane at STP.

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