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The excess reactant in the combustion of 211 g of methane in an open atmosphere is oxygen, O₂
<h3>What is a combustion reaction? </h3>
A combustion reaction is a reaction in which a substance burns in air (oxygen) to produce carbon dioxide and water.
<h3>Combustion of methane </h3>
Methane combust in air (oxygen) to produce carbon dioxide and water according to the following equation:
CH₄ + 2O₂ —> CO₂ + 2H₂O
From the question given above, we were told that 211 g of methane combust in an open atmosphere. This simply means that the amount of oxygen will be very high.
Therefore, we can conclude that oxygen will be the excess reactant since the reaction is carried out in an open atmosphere.
Learn more about stoichiometry:
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