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What is the volume of oxygen required to burn completely 45cm3 of methane at stp ?
Accepted Answer
The balanced chemical equation for the combustion of methane is:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
According to the stoichiometry of the reaction, 1 mole of methane requires 2 moles of oxygen for complete combustion.
At STP, 1 mole of any gas occupies a volume of 22.4 L.
Therefore, the volume of oxygen required to burn 45cm3 of methane at STP is:
Volume of oxygen = (45cm3 / 22400 cm3/mol) * (2 mol O2 / 1 mol CH4) * (1000 cm3 / 1 L)
= 40 cm3
Hence, the volume of oxygen required to burn 45cm3 of methane completely at STP is 40 cm3.
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