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Answer :
Final answer:
The volume of oxygen needed to combust 4.7 L of methane at STP is 9.4 L, following the 1:2 molar ratio from the balanced chemical equation for the combustion of methane. Therefore, The correct answer is option b. 9.4 L.
Explanation:
The question is asking about the stoichiometry of the combustion reaction of methane with oxygen. According to the balanced chemical equation CH4 + 2O2 → CO2 + 2H2O, one mole of methane reacts with two moles of oxygen. Since gas volumes are directly proportional to moles at Standard Temperature and Pressure (STP), we can use the volumes in the same ratio.
For the combustion of 4.7 L of methane, the volume of oxygen needed is twice that of methane because of the 1:2 molar ratio. So, the required volume of oxygen at STP would be 4.7 L × 2 = 9.4 L. Therefore, the correct answer is 9.4 L, which matches option b).
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