Answer the following question based on the chemical equation below. 2 C2H6 (g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)A student mixes 3.85 moles of C2H6 with 15.0 moles of O2 during an experiment.What is the maximum amount (in grams) of CO2 that can be produced during this experiment?

A chemical equation is given as follows:In the above reaction, 2 moles of C2H6 react with 7 moles of O2 to produce 4 moles of CO2 and 6 moles of H2O.A reaction in which a substance like hydrocarbon reacts with excess oxygen to produce carbon dioxide gas and water is called a combustion reaction.So the reaction between ethane, C2H6, and oxygen, O2 is a combustion reaction.The number of moles of C2H6 is given as .The number of moles of O2 is given as .We have to determine the maximum amount (in grams) of CO2 that can be produced.The moles to stoichiometric coefficient ratio for C2H6 can be calculated as follows:The moles to stoichiometric coefficient ratio for O2 can be calculated as follows:Since the moles to the stoichiometric coefficient ratio for C2H6 is less than the moles to the stoichiometric coefficient ratio for O2 , so C2H6 is the limiting reagent.A limiting reagent is one of the reactants that gets completely consumed during the reaction. 2 moles of C2H6 produce 4 moles of CO2, so the number of moles of CO2 produced by the 3.85 moles of C2H6 can be calculated as follows:The formula to calculate mass using moles and molar mass is given as follows:The molar mass of CO2 is .Substitute for the moles of CO2 and for the molar mass of CO2 in the above formula to calculate the mass of CO2 produced.Therefore, CO2 can be produced during the experiment.The amount of CO2 produced is calculated as .