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.<br /><br />Given information:<br />- 56 grams of $C_{2}H_{4}$<br />- 96 grams of $O_{2}$<br />- Balanced chemical equation: $C_{2}H_{4} + 3O_{2} \rightarrow 2CO_{2} + 2H_{2}O$<br /><br />Step 1: Calculate the molar mass of each reactant.<br />- Molar mass of $C_{2}H_{4}$ = 28.05 g/mol<br />- Molar mass of $O_{2}$ = 32.00 g/mol<br /><br />Step 2: Calculate the number of moles of each reactant.<br />- Moles of $C_{2}H_{4}$ = 56 g / 28.05 g/mol = 2.00 mol<br />- Moles of $O_{2}$ = 96 g / 32.00 g/mol = 3.00 mol<br /><br />Step 3: Compare the number of moles of each reactant to the stoichiometric ratio in the balanced chemical equation.<br />- The balanced chemical equation shows that 1 mole of $C_{2}H_{4}$ reacts with 3 moles of $O_{2}$ to produce 2 moles of $CO_{2}$.<br />- The ratio of $C_{2}H_{4}$ to $O_{2}$ in the balanced chemical equation is 1:3.<br />- The ratio of the number of moles of $C_{2}H_{4}$ to $O_{2}$ is 2.00 mol : 3.00 mol = 2:3.<br /><br />Since the ratio of the number of moles of $C_{2}H_{4}$ to $O_{2}$ is less than the stoichiometric ratio of 1:3, $C_{2}H_{4}$ is the limiting reagent.<br /><br />Therefore, the limiting reagent is $C_{2}H_{4}$.