Limiting Reagents & Percent Yield Chapter 9 Notes Part III

What are limiting reagents? Up until now, we have assumed that all reactants are used up in a reaction. In actuality, usually one chemical runs out first, stopping the reaction—that chemical is called the limiting reagent.

For Example: How many sandwiches can you make from a full jar of peanut butter, a full jar of jelly and 4 slices of bread?

For Example: How many sandwiches can you make from a full jar of peanut butter, a full jar of jelly and 4 slices of bread?

For Example: How many sandwiches can you make from a full jar of peanut butter, a full jar of jelly and 4 slices of bread? +

For Example: How many sandwiches can you make from a full jar of peanut butter, a full jar of jelly and 4 slices of bread? +

For Example: How many sandwiches can you make from a full jar of peanut butter, a full jar of jelly and 4 slices of bread? ++

For Example: How many sandwiches can you make from a full jar of peanut butter, a full jar of jelly and 4 slices of bread? ++

For Example: How many sandwiches can you make from a full jar of peanut butter, a full jar of jelly and 4 slices of bread? ++=

For Example: How many sandwiches can you make from a full jar of peanut butter, a full jar of jelly and 4 slices of bread? ++=

The limiting reagent limits or determines the amount of product that can be formed in a reaction. On the flip side, the excess reagent is the one not completely used up.

How do you find what the limiting reagent is? To find the limiting reagent, use dimensional analysis to determine how much of the product will be formed by both reactants. The smaller of the two results will indicate which reactant is the limiting reagent.

Here’s an example: In the reaction: Mg + HCl  MgCl 2 + H 2 you combine 35.5 grams of magnesium with 28.1 grams of hydrochloric acid. Which one is the limiting reagent, and how much magnesium chloride is produced?

Mg + 2 HCl  MgCl 2 + H 2

35.5 g Mg x 1 mole Mg x 1 mole MgCl 2 x 95.3 g MgCl 2 = 139 g MgCl g Mg 1 mole Mg 1 mole MgCl g Mg 1 mole Mg 1 mole MgCl 2

Mg + 2 HCl  MgCl 2 + H g Mg x 1 mole Mg x 1 mole MgCl 2 x 95.3 g MgCl 2 = 139 g MgCl g Mg 1 mole Mg 1 mole MgCl g Mg 1 mole Mg 1 mole MgCl g HCl x 1 mole HCl x 1 mole MgCl 2 x 95.3 g MgCl 2 = 36.6 g MgCl g HCl 2 moles HCl 1 mole MgCl g HCl 2 moles HCl 1 mole MgCl 2

Mg + 2 HCl  MgCl 2 + H g Mg x 1 mole Mg x 1 mole MgCl 2 x 95.3 g MgCl 2 = 139 g MgCl g Mg 1 mole Mg 1 mole MgCl g Mg 1 mole Mg 1 mole MgCl g HCl x 1 mole HCl x 1 mole MgCl 2 x 95.3 g MgCl 2 = 36.6 g MgCl g HCl 2 moles HCl 1 mole MgCl g HCl 2 moles HCl 1 mole MgCl 2 HCl is the limiting reagent and 36.6 g of MgCl 2 are produced.

Percent Yield In the real world, you never form 100% of the products that you think you should based on theoretical calculations. By comparing how much of a product forms to how much should form based on stoichiometry, you can find the percent yield.

Percent Yield %Yield= actual yield x 100 theoretical yield

Practice Problem In the reaction of nitrogen gas and hydrogen gas producing ammonia (NH 3 ), 250.2g H 2 produces g NH 3. What is the % yield of this reaction?

N H 2 → 2 NH 3 First find out how much NH 3 should be produced:

N H 2 → 2 NH 3 First find out how much NH 3 should be produced: g H 2 x 1 mole H 2 x 2 moles NH 3 x 17 g NH 3 = 2836 g NH 3 1 g H 2 3 moles H 2 1 mole NH 3

N H 2 → 2 NH 3 First find out how much NH 3 should be produced: g H 2 x 1 mole H 2 x 2 moles NH 3 x 17 g NH 3 = 2836 g NH 3 1 g H 2 3 moles H 2 1 mole NH 3 Then calculate the percent yield:

N H 2 → 2 NH 3 First find out how much NH 3 should be produced: g H 2 x 1 mole H 2 x 2 moles NH 3 x 17 g NH 3 = 2836 g NH 3 1 g H 2 3 moles H 2 1 mole NH 3 Then calculate the percent yield: % Yield = Actual x 100 % Yield = g x 100 Theor g % Yield = 38.6 %

N H 2 → 2 NH 3 First find out how much NH 3 should be produced: g H 2 x 1 mole H 2 x 2 moles NH 3 x 17 g NH 3 = 2836 g NH 3 1 g H 2 3 moles H 2 1 mole NH 3 Then calculate the percent yield: % Yield = Actual x 100 % Yield = g x 100 Theor g % Yield = 38.6 %

Reaction #1 K 2 CrO 4 + Pb(NO 3 ) 2  KNO 3 + PbCrO 4 If 0.947g K 2 CrO 4 are combined with 0.331g Pb(NO 3 ) 2 what will be the limiting reagent, and how much lead chromate is produced?

K 2 CrO 4 + Pb(NO 3 ) 2  2 KNO 3 + PbCrO g K 2 CrO 4 x 1 mole K 2 CrO 4 x 1 mole PbCrO 4 x g PbCrO 4 = 1.97 g PbCrO g K 2 CrO 4 1 mole K 2 CrO 4 1 mole PbCrO g Pb(NO 3 ) 2 x 1 mole Pb(NO 3 ) 2 x 0.331g Pb(NO 3 ) 2 x 1 mole Pb(NO 3 ) 2 x 1 mole PbCrO 4 x g PbCrO 4 = g PbCrO 4 Pb(NO 3 ) 2 1 mole Pb(NO 3 ) g Pb(NO 3 ) 2 1 mole Pb(NO 3 ) 2 1 mole PbCrO 4 Pb(NO 3 ) 2 is the limiting reagent and g PbCrO 4 is produced g PbCrO 4 is produced.

Reaction #2 C 3 H 8 O + O 2  CO 2 + H 2 O If 3.78 L of O 2 are combined with 7.9 g of C 3 H 8 O, what is the limiting reagent, and how much water is produced?

2 C 3 H 8 O + 9 O 2  6 CO H 2 O 3.78 L of O 2 x 1 mole O 2 x 8 moles H 2 O x 18 g H 2 O = 2.7 g H 2 O 22.4 L O 2 9 moles O 2 1 mole H 2 O 22.4 L O 2 9 moles O 2 1 mole H 2 O 7.9 g of C 3 H 8 O x 1 mole C 3 H 8 O x 8 moles H 2 O x 18 g H 2 O = 9.48 g H 2 O 60 g C 3 H 8 O 2 moles C 3 H 8 O 1 mole H 2 O 60 g C 3 H 8 O 2 moles C 3 H 8 O 1 mole H 2 O O 2 is the limiting reagent and 2.7 g of H 2 O are produced.

Reaction #3 K 2 CrO 4 + AgNO 3  KNO 3 + Ag 2 CrO 4 If 1.17g K 2 CrO 4 are combined with 1.23g AgNO 3, what is the limiting reagent and how much Ag 2 CrO 4 is produced?

K 2 CrO AgNO 3  2 KNO 3 + Ag 2 CrO g K 2 CrO 4 x 1 mole K 2 CrO 4 x 1 mole Ag 2 CrO 4 x g Ag 2 CrO 4 = 1.99 g Ag 2 CrO g K 2 CrO 4 1 mole K 2 CrO 4 1 mole Ag 2 CrO g K 2 CrO 4 1 mole K 2 CrO 4 1 mole Ag 2 CrO g AgNO 3 x 1 mole AgNO 3 x 1 mole Ag 2 CrO 4 x g Ag 2 CrO 4 = 1.20 g Ag 2 CrO g AgNO 3 2 moles AgNO 3 1 mole Ag 2 CrO g AgNO 3 2 moles AgNO 3 1 mole Ag 2 CrO 4 AgNO 3 is the limiting reagent and AgNO 3 is the limiting reagent and 1.20 g Ag 2 CrO 4 are produced g Ag 2 CrO 4 are produced.

Reaction #4 Zn + HCl  ZnCl 2 + H 2 What is the limiting reagent and how many liters of hydrogen are produced if 15.5 grams of zinc are combined with 26.3 grams of hydrochloric acid?

Zn + 2 HCl  ZnCl 2 + H g Zn x 1 mole Zn x 1 mole H 2 x 22.4 L H 2 = 5.31 L H g Zn 1 mole Zn 1 mole H g HCl x 1 mole HCl x 1 mole H 2 x 22.4 L H 2 = 8.07 L H g HCl 2 moles HCl 1 mole H 2 Zn is the limiting reagent and 5.31 Liters of H 2 are produced.

Reaction #5 Na + H 2 O  NaOH + H 2 If 2.1g of sodium is added to 3.8g of water, what is the limiting reagent, and how many grams of sodium hydroxide are produced?

2 Na + 2 H 2 O  2 NaOH + H g Na x 1 mole Na x 2 moles NaOH x 40 g NaOH = 3.7 g NaOH 23 g Na 2 moles Na 1 mole NaOH 3.8 g H 2 O x 1 mole H 2 O x 2 moles NaOH x 40 g NaOH = 8.4 g NaOH 18 g H 2 O 2 moles H 2 O 1 mole NaOH Na is the limiting reagent and 3.7 g NaOH are produced.

In Review: What is a limiting reagent? What is the excess reagent? How does a limiting reagent control the amount of product?