Limiting Reagent u If you are given one dozen loaves of bread, a gallon of mustard and three pieces...

Limiting Reagent

If you are given one dozen loaves of bread, a gallon of mustard and three pieces of salami, how many salami sandwiches can you make?

Limiting Reagent

The limiting reagent is the reactant you run out of first.

The excess reagent is the one you have left over.

The limiting reagent determines how much product you can make.

Determining the Limiting Reactant

Do two stoichiometry problems. The one that makes the least product

is the limiting reagent.

Example

Copper reacts with sulfur to form copper (I) sulfide. If 10.6 g of copper reacts with 3.83 g sulfur, how much product will be formed?

2 Cu + S Cu2S

10.6 g Cu 63.5 g Cu

1 mol Cu

2 mol Cu

1 mol Cu2S 1 mol Cu2S

159.1 g Cu2S

= 13.3 g Cu2S

Example

2 Cu + S Cu2S

3.83 g S 32.1 g S

1 mol S1 mol S

1 mol Cu2S 1 mol Cu2S

159.1 g Cu2S

= 19.0 g Cu2S

Example

2 Cu + S Cu2S

Example

10.6 grams of copper resulted in 13.3 grams of copper (I) sulfide.

3.83 grams of sulfur resulted in 19.0 grams of copper (I) sulfide.

2 Cu + S Cu2S

Cu is the limiting reactant!

Example

Using 10.6 grams of copper resulted in the smaller amount of copper (I) sulfide formed. Therefore

2 Cu + S Cu2S

Example

Considering the limiting reactant, 13.3 grams of copper (I) sulfide are formed.

Example

Determine the mass of P4O10 formed if 25.0 g of P4 and 50.0 g of O2 are combined. Which reactant is the limiting reactant?

Write the balanced equation P4 + 5 O2 P4O10

Example

P4 + 5 O2 P4O10

25.0 g P4

124.0 P4

1 mol P4

1 mol P4

1 mol P4O10

1 mol P4O10

284.0 g P4O10

= 57.3 g P4O10

Example

P4 + 5 O2 P4O10

50.0 g O2

32.0 O2

1 mol O2

5 mol O2

1 mol P4O10

1 mol P4O10

284.0 g P4O10

= 88.8 g P4O10

Example

Example

P4 is the limiting reactant, and 57.3 grams of P4O10 are formed.

Problem

The reaction between solid sodium and iron (III) oxide is one in a series of reactions that inflates an automobile airbag. If 100. g of Na and 100. g of Fe2O3 are used in the reaction, determine the limiting reactant and the mass of solid iron produced. Fe2O3 limits, 69.9 g

of Fe produced

Problem

How many moles of cesium xenon heptafluoride (CsXeF7) can be produced from the reaction of 12.5 mol of cesium fluoride with 10.0 mol of xenon hexafluoride?

10.0 mol

Problem

The reaction of chlorine gas with solid phosphorus (P4) produces solid PCl5. When 22.0 g of chlorine reacts with 25.5 g of P4, which reactant limits the amount of product?

Cl2 is limiting.

How Much Excess Reagent?

Convert the limiting reagent into the excess reagent.

Subtract that from the amount of excess you started with.

Example

Mg(s) + 2 HCl(g) MgCl2(s) + H2(g) If 10.1 mol of magnesium and

4.87 mol of HCl gas are reacted, how many moles excess reagent remain?

Example

Mg(s) + 2 HCl(g) MgCl2(s) + H2(g) First, determine which reactant is in

excess.

Mg(s) + 2 HCl(g) MgCl2(s) + H2(g)

10.1 mol Mg

1 mol Mg

1 mol MgCl2 = 10.1 mol MgCl2

Example

Mg(s) + 2 HCl(g) MgCl2(s) + H2(g)

4.87 mol HCl

2 mol HCl

1 mol MgCl2 = 2.44 mol MgCl2

Example

Example

Mg is in excess. Now determine how much

excess! Convert the limiting reactant to

excess reactant.

Mg(s) + 2 HCl(g) MgCl2(s) + H2(g)

4.87 mol HCl

2 mol HCl

1 mol Mg = 2.44 mol Mg

Example

Example

Now take the excess reactant and subtract the liming converted into excess.

10.1 mol Mg – 2.44 mol Mg =

7.66 mol of Mg are in excess

Example

If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much excess reagent will remain?

Example

Write the balanced equation. If 10.3 g of aluminum are reacted

with 51.7 g of CuSO4 how much excess reagent will remain?

2Al + 3CuSO4 Al2(SO4)3 + 3Cu

Example

2Al + 3CuSO4 Al2(SO4)3 + 3Cu First, determine which reactant is in excess.

2Al + 3CuSO4 Al2(SO4)3 + 3Cu

10.3 g Al

27.0 g Al

1 mol Al

2 mol Al

3 mol Cu

= 0.572 mol Cu

Example

2Al + 3CuSO4 Al2(SO4)3 + 3Cu

51.7 g CuSO4

159.6 g CuSO4

1 mol CuSO4

3 mol CuSO4

3 mol Cu

= 0.324 mol Cu

Example

Example

Al is in excess. Now determine how much

excess! Convert the limiting reactant to

excess reactant.

2Al + 3CuSO4 Al2(SO4)3 + 3Cu

51.7 g CuSO4

159.6 CuSO4

1 mol CuSO4

3 mol CuSO4

2 mol Al

1 mol Al

27.0 g Al

= 5.83 g Al

Example

Example

Now take the excess reactant and subtract the liming converted into excess.

10.3 g Al – 5.83 g Al =

4.47 g of Al are in excess

Problem

Zn + 2MnO2 + H2O Zn(OH)2 + Mn2O3

If 25.0 g of zinc and 39.0 g of manganese dioxide are used, how much excess reagent will remain?

(Answer: 10.3 g Zn)

Problem

Lithium reacts spontaneously with bromine to produce lithium bromide. If 3.40 g of lithium react with 32.0 g of bromine, how much excess reactant remains?

(Answer: 0.64 g Li)

Problem

CH4 + 3Cl2 CH3Cl + 3HCl If 50.0 g of methane reacts with

9.67 moles of chlorine gas, how mass of excess reagent will remain?

(Answer: 20.6 g of Cl2)

Yield

The yield is the amount of product made in a chemical reaction.

There are three types of yield that can be determined.

Yield Actual yield - what you get in the lab

when the chemicals are mixed

Theoretical yield - what the balanced equation tells you you should make

Actual Percent yieldPercent yield = x 100 %

Theoretical

Percent Yield

Percent yield tells us how “efficient” a reaction is.

Percent yield cannot be larger than 100 %.

Example

6.78 g of copper are produced when 3.92 g of Al are reacted with excess copper (II) sulfate.

2Al + 3 CuSO4 Al2(SO4)3 + 3Cu

What is the actual yield of copper?

(Answer: 6.78 g)

Example

6.78 g of copper are produced when 3.92 g of Al are reacted with excess copper (II) sulfate.

2Al + 3 CuSO4 Al2(SO4)3 + 3Cu

What is the theoretical yield?

2Al + 3 CuSO4 Al2(SO4)3 + 3Cu

3.92 g Al

27.0 Al

1 mol Al

2 mol Al

3 mol Cu

1 mol Cu

63.5 g Cu

= 13.8 g Cu

Example

Example

6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.

2Al + 3 CuSO4 Al2(SO4)3 + 3Cu

What is the percent yield?

Yield

Actual Percent yieldPercent yield = x 100 %

Theoretical

6.78 g Percent yieldPercent yield = x 100 %

13.8 g

= 49.1%

Problem

58.2 g of zinc are produced when 14.7 g of Li are reacted with excess zinc sulfate.

2Li + ZnSO4 Li2SO4 + Zn

What is the actual yield of zinc?

(Answer: 58.2 g)

Problem

58.2 g of zinc are produced when 14.7 g of Li are reacted with excess zinc sulfate.

2Li + ZnSO4 Li2SO4 + Zn

What is the theoretical yield of zinc?

(Answer: 69.7 g)

Problem

58.2 g of zinc are produced when 14.7 g of Li are reacted with excess zinc sulfate.

2Li + ZnSO4 Li2SO4 + Zn

What is the percent yield of zinc?

(Answer: 83.5%)