What is the excess reactant in the combustion of 211 g of methane in an open atmosphere?

Name: What is the excess reactant in the combustion of 211 g of methane in an open atmosphere?
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In the combustion of methane (CH4) in an open atmosphere, the reaction is CH4 + 2O2 -> CO2 + 2H2O. Given that oxygen (O2) is abundant in the open atmosphere, it is not the limiting reactant. Methane, being a specific quantity, is entirely consumed assuming complete combustion, leaving oxygen as the excess reactant.

What Is the Excess Reactant in the Combustion of 211