Question

Calculate the volume of oxygen (in litres) that is needed for the total combustion of 28 L of methane (STP).

pls. solve it briefly with solution​

Answer 1

Answer:

Given volume of methane =0.25dm³

=0.25L

Combustion of methane (CH4) -

CH4 + 2O2 ⟶ CO2 + 2H2O

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

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

= (44.8/22.4)(0.25)

=0.5 L

Hence, 0.5 L of oxygen required for the complete combustion of 0.25dm³ of methane.

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