STOICHIOMETRY IS THE PART OF CHEMISTRY THAT DEALS WITH THE AMOUNTS OF

SUBSTANCES INVOLVED IN CHEMICAL REACTIONS.

 HENCE, IT IS THE STUDY OF THE

QUANTITATIVE RELATIONSHIPS IN CHEMICAL REACTIONS.

 A BALANCED CHEMICAL EQUATION INDICATES

THE RELATIVE NUMBER OF MOLES OR “PARTICLES” INVOLVED IN A CHEMICAL

REACTION.

Stoichiometry

H2O(g

)

NO(g) O2(g

)

NH3(g

) ++4 645

4 molecules5 molecules 4 molecules 6 molecules

4 moles 5 moles 4 moles 6 moles

68.16 g 160.00 g 120.04 g 108.12 gRecall the definition of a mole:

We can expand this definition to include gases.

Early chemists primarily worked with gases, and devised many laws to explain the behavior(s) of these gases.

g-formula mass = 1 mole = 6.022 x 1023 particles

In 1811 Avogadro postulated his own gas law.

He said that, “Equal volumes of gases contain equal

numbers of molecules.”

Thus, the volume is directly proportional to the number of particles of the number of moles.

g-formula mass = 1 mole = 6.022 x 1023 particles

= 22.4 dm3 at STP**Standard Temperature and Pressure

0C and 1.0 atm

H2O(g

)

NO(g) O2(g

)

NH3(g

) ++4 645

4 liters 5 liters 4 liters 6 liters

By writing balanced chemical equations and incorporating mole conversions, you can calculate the amount of reactants needed or the amount of products produced. This is due to the fact that the equation indicates the number of moles of reactants and products.

TO SOLVE:

1. Write the balanced equation.

2. Find the number of moles of the given substance.

3. Inspect and use the balanced equation to determine the ratio of moles of given substance to moles of “wanted” substance.

4. Convert the amount of “wanted” substance into the desired units.

THE TYPICAL FORMAT:

(given) grams 1 mole (given) moles of wanted g-formula mass

(wanted)

g-formula mass (given) moles of given 1 mole (wanted)Step#2

Step#3from equation

Step#4

Example: Sulfuric acid is neutralized by the addition of sodium hydroxide.

How many grams of water are produced when 15. grams of sodium

hydroxide react?TO SOLVE:1. Write the balanced equation.

2. Find the number of moles of the given substance.

3. Inspect and use the balanced equation to determine the ratio of moles of given substance to moles of “wanted” substance.

4. Convert the amount of “wanted” substance into the desired units.

H2SO4(aq)

+ NaOH(aq)

Na2SO4(aq) + HOH(l)+ H2O(l)2 215 g

??? g

15 g NaOH 1 mole NaOH

40.00 g NaOH

2 mole H2O

2 mole NaOH

18.02 g H2O

1 mole H2O

= 6.8 g H2O

Example: How many moles of NaOH are required to neutralize 4.2 moles of H2SO4?

TO SOLVE:1. Write the balanced equation.

2. Find the number of moles of the given substance.

3. Inspect and use the balanced equation to determine the ratio of moles of given substance to moles of “wanted” substance.

4. Convert the amount of “wanted” substance into the desired units.

H2SO4(aq)

+ NaOH(aq)

Na2SO4(aq) + H2O(l)2 2??? moles4.2 moles

4.2 moles H2SO4

2 moles NaOH

1 mole H2SO4

= 8.4 moles NaOH

Example: How many grams of Na2SO4 are produced from 4.2 moles of H2SO4?

TO SOLVE:1. Write the balanced equation.

2. Find the number of moles of the given substance.

3. Inspect and use the balanced equation to determine the ratio of moles of given substance to moles of “wanted” substance.

4. Convert the amount of “wanted” substance into the desired units.

H2SO4(aq)

+ NaOH(aq)

Na2SO4(aq) + H2O(l)2 24.2 moles ??? g

4.2 moles H2SO4

1 mole Na2SO4

1 mole H2SO4

142.04 gNa2SO4

1 mole Na2SO4= 596.568 g Na2SO4= 6.0 x 102 g Na2SO4

Limiting ReagentsSo far you have worked stoichiometry problems in which the given quantity of a reactant is consumed completely and we use that quantity to figure out how much product is formed. However, in most instances you mix different amounts of various reactants and not all completely react. Lets look at the following hypothetical reaction:A +

BC

Start: 5 mol 3 mol 0 molEnd: 2 mol 0 mol 3 mol

As you can clearly see, all three moles of B were consumed but since you had an excess of A, 2 moles were left over.

The reagent that is consumed 100% is referred to as the limiting reagent or reactant (LR).

The LR allows you to calculate (just like you learned previously) the amount of product formed and how much of the other reactant was consumed.

This is a lot simpler to understand with an example.

Limiting reactant analogies….

For example, bicycles require one frame and two wheels. If you have 20 wheels but only 5 frames, it is clear that the number of frames will determine how many bicycles can be made.

TO SOLVE LIMITING REACTANT (REAGENT) PROBLEMS:

1. Write the balanced equation for the reaction.

2. Calculate the number of moles of each reactant.

3. Convert one of the reactants into the other. This tells you how much of each you need to complete the reaction.

4. Compare the moles calculated in step #2 to the amounts in #3. The one you have less of is the limiting reagent.

5. Use the limiting reagent as the given, and calculate the desired quantity.

EXAMPLE: The Haber Process is used to convert nitrogen

and hydrogen gases to ammonia. If 20. g N2 and 10. g H2 are mixed, what is the limiting reagent?

N2(g) + H2(g) NH3(g)3 210. g20. g

1. Write the balanced equation for the reaction.

2. Calculate the number of moles of each reactant.

20. g N2 1 mole N2

28.02 g N2

10. g H2 1 mole H2

2.02 g H2

= 0.71 moles N2(g)= 5.0 moles H2(g)

3. Convert one of the reactants into the other. This tells you how much of each you need to complete the reaction.

4. Compare the given amounts to the answer in #3. The one you have less of is the limiting reagent.

• You have 5 moles of H2…..more than enough…

N2 is the limiting reagent. There is an excess of H2, After the reaction has completed there will be no N2 left, yet there will still be extra H2.

= .71 N2 given and 2.1 moles need to react with it

.713775874375 moles N2

3 mole H2

1 mole N2

= 2.1 moles H2 needed

EXAMPLE: How much NH3 is produced?

Use the L R (N2 in this problem) to determine amounts of products…..

From the last example, we know

N2(g) + H2(g) NH3(g)3 2

.713775874375 moles N2

2 mole NH3

1 mole N2

17.04 g NH3

1 mole NH3

= 24 g NH3

EXAMPLE: How much excess is there?

Use the L R (N2 in this problem) to determine how much of the other reactant (H2 in this problem) is left over.

From the last example, we know

N2(g) + H2(g) NH3(g)3 2

.713775874375 moles N2

3 mole H2

1 mole N2

2.02 gH2

1 mole H2

= 4.32548179871 g H2

reacted 10. g H2 to start

- 4.325481 g H2 reacted

5.674518 g in excess = 6 g H2 left over

EXAMPLE: How many grams of aluminum sulfide can form

from the reaction of 9.00 g of aluminum with

8.00 g of sulfur?Al(s) + S(g) Al2S3(s)3

8.00 g9.00 g

9.00 g Al 1 mole Al

26.98 g Al

8.00 g S 1 mole S

32.07 g S

= 0.334 moles Al(S)= 0.249 moles S(s)

2??? g

There is more than enough aluminum

0.249 moles S 2 moles Al

3 moles S

Sulfur is the limiting reactant

= 0.166 moles Al needed

Use the L R (S in this problem) to determine amounts of products…..

0.249454318678 moles S

1 mole Al2S3

3 mole S

150.17 g Al2S3

1 mole Al2S3

= 12.5 g Al2S3 produced

Al(s) + S(g) Al2S3(s)38.00 g9.00 g

2??? g

EXAMPLE: How many grams of N2F4 can theoretically be

obtained from 4.00 g of NH3 and 14.0 g of F2 according to:

NH3 + F2 N2F4514.00 g4.00 g

4.00 g NH3 1 mole NH3

17.04 g NH3

14.00 g F2 1 mole F2

38.00 g F2

= 0.235 moles NH3= 0.3684 moles F2

2??? g

0.235 moles NH3

5 moles F2

2 moles NH3

= .5875 moles F2 needed

Fluorine is the limiting reactant

+ HF6

There is NOT ENOUGH fluorine

Use the L R (F2 in this problem) to determine amounts of products…..

0.368421052632 moles F2

1 mole N2F4

5 mole F2

104.02 g N2F4

1 mole N2F4

= 7.665 g N2F4 produced

NH3 + F2 N2F4514.00 g4.00 g

2??? g

+ HF6

Lets say you carry out the reaction described above in the lab and you only obtain 4.80 g N2F4. What happened????

The quantities we calculate in mass-mass (Stoichiometry) problems are THEORETICAL AMOUNTS; that is, the maximum possible yield.

In reality, the actual amount of products is much less due to human error, mechanical error, and possible side reactions.

By comparing the actual versus theoretical, scientists calculate the percentage yield.% Yield =

Actual

Theoretical

 NOTE: % Error + % Yield = 100%

4.80 g N2F4 actual x 100 %7.665 g N2F4

theoretical

From the last problem

% Yield =Actual

X 100%Theoretical

= 62.6% N2F4

yielded