What is the limiting reagent when 56 grams of C_(2)H_(4) react-with 96 grams of O_(2) to produce 88 grams of CO_(2) C_(2)H_(4)+3O_(2)arrow 2CO_(2)+2H_(2)O CO_(2) O_(2) There is not a limiting reagent because the reaction involves stoichiometric amounts of each substance. C_(2)H_(4)

To determine the limiting reagent, we need to compare the amount of each reactant to the amount required by the balanced chemical equation.Given information:- 56 grams of - 96 grams of - Balanced chemical equation: Step 1: Calculate the molar mass of each reactant.- Molar mass of = 28.05 g/mol- Molar mass of = 32.00 g/molStep 2: Calculate the number of moles of each reactant.- Moles of = 56 g / 28.05 g/mol = 2.00 mol- Moles of = 96 g / 32.00 g/mol = 3.00 molStep 3: Compare the number of moles of each reactant to the stoichiometric ratio in the balanced chemical equation.- The balanced chemical equation shows that 1 mole of reacts with 3 moles of to produce 2 moles of .- The ratio of to in the balanced chemical equation is 1:3.- The ratio of the number of moles of to is 2.00 mol : 3.00 mol = 2:3.Since the ratio of the number of moles of to is less than the stoichiometric ratio of 1:3, is the limiting reagent.Therefore, the limiting reagent is .