Limiting Reagents EQ: What happens in a chemical reaction, if there is an insufficient amount of one reactant?

Limiting Reagents Limiting Reagent -- the reactant that is completely used up first in the chemical reaction. Excess Reagent(s) -- the reactant(s) present in quantities greater than necessary to react with the quantity of the limiting reagent. (not completely used up)

Limiting Reagent The concept of limiting reagent is analogous to the relationship between men and women in a dance at a club. If there are 14 men and only 9 women, how many female/male pairs can compete? Only 9 female/male pairs can compete. Five men will be left without partners. Who is Limiting? The number of women limits the number of men that can compete Who is in excess? Men are in excess

Example # 1 Consider the formation of nitrogen dioxide (NO2) from nitric oxide (NO) and oxygen (O2) 2 NO(g) + O2(g) -----> 2 NO2(g) Suppose initially we have 8 moles of NO and 7 moles of O2. One way to determine the limiting reagent is to calculate the number of moles of NO2 obtained based on the initial quantities of NO and O2. Remember the limiting reagent will yield the smaller amount of the product.

Example #1 Calculate the number of moles of product obtained from each reactant. The one that yields the smallest amount of product is the limiting reagent.

2 NO(g) + O2(g) -----> 2 NO2(g) Starting with 8 moles of NO, we find the number of moles of NO2 produced is = Now, starting with 7 moles of O2, the number of moles of NO2 produced is 8 mol NO 2 mol NO2 8 mol NO2 2 mol NO 7 mol O2 2 mol NO2 14 mol NO2 1 mol O2

Example#1 Because NO yields the smaller amount of NO2, it must be the limiting reagent. Therefore, O2 is the excess reagent.

Example # 2 The depletion of ozone (O3) in the stratosphere has been a matter of great concern among scientists in recent years. It is believed that ozone can react with nitric oxide (NO) that is discharged from the high-altitude jet plane, the SST. The reaction is O3 + NO -----> O2 + NO2

Example # 2 - continued If 0.710 g of O3 reacts with 0.670 g of NO, how many grams of NO2 will be produced? Which compound is the limiting reagent? Calculate the number of grams of the excess reagent remaining at the end of the reaction?

This equation as it is written is balanced. Example # 2 - continued First we need to make sure the equation is balanced: O3 + NO -----> O2 + NO2 This equation as it is written is balanced.

Example # 2 - continued a.) = 0.680 g NO2 = 1.027 g NO2 0.710 g O3 1 mole O3 1 mole NO2 46.01 g NO2 48.00 g O3 1 mole O3 1 mole NO2 0.670 g NO 1 mole NO 1 mol NO2 46.01g NO2 30.01 g NO 1 mol NO 1 mol NO2

Example # 2 - continued b) Because O3 yields the smallest amount of NO2, it is the limiting reagent. Then, NO must be the excess reagent. 1O3 + 1NO -----> 1O2 +1NO2 Because there is a 1 to 1 mole ratio between NO and NO2, then the amount of excess reagent left after the reaction is complete, assuming all of the O3 reacted, is: = 0.444 g NO Therefore, 0.670 g NO – 0.444 g NO = 0.226 g NO remaining 0.710 g O3 1 mol O3 1 mol NO 30.01 g NO 48.00 g O3 1 mol O3 1 mol NO