stoichiometry application: we are applying concepts learned in chapters 5, 6 & 7 »...

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Stoichiometry• APPLICATION:

•We are applying concepts learned in Chapters 5, 6 & 7

»Naming/Formulas»Molar Conversions»Balancing Chemical Equations

• PURPOSE:• To understand how chemical equations are used to express ratios of reactants and products

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Stoichiometry• Stoichiometry:– From Greek: – “Stoichio” = matter or chemicals– “Metry” = measure

=The calculation of quantities in a chemical reaction.

• Kinds of quantities measured:• Particles (atoms, molecules, formula units)• Moles• Mass• Volume

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StoichiometryWhy is it useful?

• To determine the mass, volume or number of particles needed for a particular reaction.• To determine the mass, volume or number

of particles produced by a particular reaction.

Who uses stoichiometry?• Chemists,• Chemical engineers,• Engineers,• Rocket scientists,• Environmental scientists….• + criminals making illegal drugs, terrorists making bombs…

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Stoichiometry• Why are balanced chemical equations

useful?• Look at the ratios of all reactants and products

• One molecule of nitrogen reacts with three molecules of hydrogen to produce two molecules of ammonia

»1:3:2 ratio

• What happens if you have 10 molecules of N2?

1 2N2(g) + H2(g)

NH3 (g) 310

30 20

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Stoichiometry• Mass:

• All balanced chemical reactions must follow the Law of Conservation of Mass

• 1 mol N2 = 28.0 g

• 3 mol H2 = 6.0 g

• 2 mol NH3 = 34.0 g

1 2N2(g) + H2(g)

NH3 (g) 3

Reactants = 34.0 g

Products = 34.0 g

MASS IS ALWAYS CONSERVED IN CHEMICAL REACTIONS!

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Stoichiometry• Volume:

• At STP (standard temperature and pressure); 1 mol of any gas = 22.4 L

• Is volume conserved?• 1 mol N2 = 22.4 L

• 3 mol H2 = 67.2 L (3 x 22.4)

• 2 mol NH3 = 44.8 L (2 x 22.4)

1 2N2(g) + H2(g)

NH3 (g) 3

Reactants = 89.6 L

Products = 44.8 L

VOLUME IS NOT USUALLY CONSERVED!

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Stoichiometry• Moles:

• The coefficients in a balanced chemical equation represent the number of moles of that substance

• Are # of Moles conserved?• 1 mol N2

• 3 mol H2

• 2 mol NH3

1 2N2(g) + H2(g)

NH3 (g) 3

Reactants = 4 mol

Products = 2 molMOLES ARE NOT USUALLY

CONSERVED!

KEY CONCEPT: Coefficients show relative numbers of moles involved

in a reaction = the molar ratio

Stoichiometry• Moles:

• All coefficients are related to the number of moles required for a reaction.

•One mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia

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1 2N2(g) + H2(g)

NH3 (g) 3

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Stoichiometry• Mole-Mole Calculations

• Balanced chemical equations are valuable because they shoe the relative ratio of reactants to products.• You can now relate the # of moles of a reactant(s) to find the # of moles of a product(s)

1 2N2(g) + H2(g)

NH3 (g) 3

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Stoichiometry• Sample Problem 1:

–How many moles of ammonia are produced when 1.2 moles of nitrogen reacts with hydrogen?

1 2N2(g) + H2(g)

NH3 (g) 3

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Stoichiometry• Example:

–Step 1:• Look at balanced chemical equation

» 1 mol of N2 produces 2 mol NH3

–Step 2:• Use a conversion factor

»Convert mol N2 to mol NH3

1 2N2(g) + H2(g)

NH3 (g) 3

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Stoichiometry• Example: How much NH3 can be

made from 1.2 mol N2?

1 2N2(g) + H2(g)

NH3 (g) 3

1.2 mol N2

mol N2

mol NH3

= 2.4 mol NH3

2

1

Mole ratio

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– Calculate the number of moles of nitrogen required to make 9.54 mol of NH3.

1 2N2(g) + H2(g)

NH3 (g) 3

9.54 mol NH3

2 mol NH3

1 mol N2

= 4.77 mol N2

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– Calculate the number of moles of hydrogen required to make 9.54 mol of NH3.

1 2N2(g) + H2(g)

NH3 (g) 3

9.54 mol NH3

2 mol NH3

3 mol H2

= 14.3 mol H2

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Stoichiometry• Mass-Mass Calculations:–When scientist conduct a lab

experiment, they measure the amount of a particular substance using grams…not moles

– Remember, you can convert the mass of a substance into moles (1-step conversion)

Mass (grams) A moles A

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Stoichiometry• Mass-Mass Calculations:– Additionally, it is possible to convert

between masses of reactants and products.

– The mole ratio (obtained from the balanced chemical equation) is the key to convert from the mass of substance A to mass of substance B

– Remember, the balanced chemical equation provides the mole ratio!

mass A moles A moles B mass B

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Stoichiometry• Mass-Mass Example #1:

– Calculate the number of grams of NH3 produced by the reaction of 6.80 g of hydrogen with an excess of nitrogen. Use the balanced equation above.

1 2N2(g) + H2(g)

NH3 (g) 3

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– Step 1: What are we calculating?• Grams H2 grams NH3

– Step 2: What is the path we need to take?

1 2N2(g) + H2(g)

NH3 (g) 3

mass H2 moles H2 moles NH3 mass NH3

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– Step 3: Complete conversions

1 2N2(g) + H2(g)

NH3 (g) 3

6.80 g H2

2.0 g H2

1 mol H2

3 mol H2

2 mol NH3

1 mol NH3

17.0 g NH3

= 38.5 g NH3

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• How many grams of nitrogen are needed to produce the 38.5 g of NH3 produced in the previous example?

1 2N2(g) + H2(g)

NH3 (g) 3

38.5 g NH3

17.0 g NH3

1 mol NH3

2 mol NH3

1 mol N2

1 mol N2

28.0 g N2

= 31.7 g N2

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• What is the total amount of mass (in grams) of the reactants in the example balanced chemical equation?

Mass of Reactants = Mass of Products

Is mass conserved in a chemical reaction???

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Mass of Reactants = Mass of Products

31.7 g N2 + ___ g H2 38.5 g NH3

ANSWER 38.5 – 31.7 = 6.8 g H2

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Stoichiometry• Other Stoichiometric Calculations:

– You can use the same steps as before to calculate the following…

»Mass-Volume»Volume-Volume»Particle-Mass

– The first step is to ALWAYS convert to moles

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Stoichiometry• Other Stoichiometric Calculations:

– Choose the appropriate road map…

particles A

mass A

volume A

particles B

mass B

volume B

moles A

moles B

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Limiting Reagent• Limiting Reagent

• Chemical equations are like recipes…» The reactants are combined to make the products

• Reactions take place based upon the mole ratios expressed in the balanced chemical equation

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• Limiting Reagent:–Limits how much product will form =

determines the amount of product that can be formed in a reaction

• Excess Reagent:–The substance(s) that is leftover because

they are in excess = there is more than enough to react with the limiting reagent

A reaction can only occur until the limiting reagent is used up!!

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Limiting Reagent• Limiting Reagent S’mores

• Graham Cracker = Gc• Marshmallow = M• Chocolate = Ch• S’more = Gc2MCh

To make one s’more…you need to have 2 graham crackers, 1 marshmallow, and 1 piece of chocolate.

2 1Gc + M + Ch

Gc2MCh 1

1

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• Limiting Reagent S’mores

• How many s’mores can you make if you have 14 (mol) graham crackers, 10 pieces of chocolate, and 8 marshmallows?

2 1Gc + M + Ch

Gc2MCh 1

1

14 mol Gc

2 mol Gc

1 mol Gc2MCh = 7.0 mol

Gc2MCh

10 mol Ch

1 mol Ch

1 mol Gc2MCh = 10.0 mol Gc2MCh

8 mol M

1 mol M

1 mol Gc2MCh = 8 mol Gc2MCh

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• Limiting Reagent S’mores

• How many s’mores can you make if you have 14 (mol) graham crackers, 10 pieces of chocolate, and 8 marshmallows?

• Graham crackers limiting reagent• THE LIMITING AGENT WILL DETERMINE

HOW MUCH PRODUCT WILL BE CREATED!

2 1Gc + M + Ch

Gc2MCh 1

1

14 mol Gc

2 mol Gc

1 mol Gc2MCh= 7.0 mol Gc2MCh

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• What would happen if 2 mol of N2 reacted with 3 mol of H2?

• NEED: Nitrogen-Hydrogen ratio 1:3• HAVE: Nitrogen-Hydrogen ratio 2:3• What reactant is limiting?• What reactant is in excess?

1 2N2(g) + H2(g)

NH3 (g) 3

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1 2N2(g) + H2(g)

NH3 (g) 3

2 mol N2

3 mol H2

1 mol N2

3 mol H2

3 mol H2

1 mol N2

6 mol H2

1 mol N2

needed

needed

But don’t have!

have!

There is not enough hydrogen for 2 mol N2, so: hydrogen is the limiting reagent!!

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• Limiting Reagent—Example #1

• What will occur when 7.20 mol of Na reacts with 3.5 mol of Cl2?

a) What is the limiting reagent?b) How many moles of NaCl are produced?c) How much of the excess reagent remains

unreacted?

2 2Na(s) + Cl2(g)

NaCl (s) 1

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a) Choose one of the reactants and convert to the amount of moles of the other reactant.

2 2Na(s) + Cl2(g)

NaCl (s) 1

7.20 mol Na

2 mol Na

1 mol Cl2= 3.6 mol Cl2

Does the value you calculated exceed the amount presented in the problem???

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b) Use the given amount (from the problem) of the limiting reagent…calculate moles of NaCl.

2 2Na(s) + Cl2(g)

NaCl (s) 1

3.5 mol Cl2

1 mol Cl2

2 mol NaCl= 7.0 mol NaCl

moles Cl2 moles NaCl

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c) The amount of excess reagent is the difference between the given amount of Na and the amount of Na needed in the reaction.

2 2Na(s) + Cl2(g)

NaCl (s) 1

3.5 mol Cl2

1 mol Cl2

2 mol Na= 7.0 mol Na

moles Cl2 moles Na

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c) The amount of excess reagent is the difference between the given amount of Na and the amount of Na needed in the reaction.

2 2Na(s) + Cl2(g)

NaCl (s) 1

7.20 mol Na – 7.00 mol Na = 0.20 mol Na in excess

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• What is the maximum number of grams of Cu2S that can be formed when 80.0 g of Cu reacts with 25.0 g of S?

a) Find the number of moles of each reactant.

2 1Cu(s) + S (g)

Cu2S (s) 1

80.0 g Cu

63.5 g Cu

1 mol Cu = 1.26 mol Cu

25.0 g S

32.1 g S

1 mol Cu = 0.779 mol S

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b) Use the balanced chemical equation to determine the limiting reagent.

2 2Cu(s) + S (g)

Cu2S (s) 1

1.26 mol Cu

2 mol Cu

1 mol S = 0.630 mol S

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• Use the balanced chemical equation to determine the limiting reagent.

2 2Cu(s) + S (g)

Cu2S (s) 1

= 0.630 mol S

Sulfur is in excess…copper is limiting

Compare the amount of sulfur calculated to the amount of sulfur presented in the problem.

0.630 mol S calculated

0.779 mol S problem

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c) Use the amount of limiting reagent to calculate the maximum amount of the product.

2 2Cu(s) + S (g)

Cu2S (s) 1

1.26 mol Cu

2 mol Cu

1 mol Cu2S = 100.2 g Cu2S1 mol Cu2S

159.1 g Cu2S

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Percent Yield• Percent Yield:

• This is a comparison of how much of a given product is created in a reaction to the ideal yield of the product in an ideal reaction.

• Theoretical Yield:» Using an equation to determine the amount of

product formed during a reaction

• Actual Yield:» The amount of a product created in an actual

laboratory experiment

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Percent Yield• Percent Yield:

• Percent yield will NEVER exceed 100%

• Percent yield can be less than 100%

• Comparison of MASS; not moles

Percent Yield

Actual Yield

Theoretical Yield= X

100%

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• Percent Yield Example #1• What is the percent yield of this reaction if

34.8 g of CaCO3 is heated to give 16.4 g of CaO?

• Determine the actual yield of CaO from the problem

» 16.4 g CaO

• Calculate the theoretical yield of CaO.

11 CaO(s) + CO2 (g)

CaCo3

(s) 1

34.8 g CaCO3

100.1 g CaCO3

1 mol CaO

= 19.5 g CaO

1 mol CaCO3

1 mol CaO

1 mol CaO

56.1 g CaO

mass CaCO3 moles CaCO3 moles CaO mass CaO

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• Percent Yield Example #1• What is the percent yield of this reaction if

34.8 g of CaCO3 is heated to give 16.4 g of CaO?

11 CaO(s) + CO2 (g)

CaCo3

(s) 1

Percent Yield = Actual Yield

Theoretical YieldX 100

Percent Yield = 16.4 g CaO

19.5 g CaOX 100

= 84.1 %

stoichiometry application: we are applying concepts learned in chapters 5, 6 & 7 »...

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