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STIOCHIOMETRY CHEMISTRY CHAPTER 9 9 .1 What is Stiochiometry  A. The study of quantitative relationships between amounts of reactants used and product formed by chemical reactions.   B. Mole-Mass relationships in chemical reactions

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9.1 What is Stiochiometry

 A. The study of quantitative relationships between amounts of reactants used and product formed by chemical reactions.

  B. Mole-Mass relationships in chemical reactions

  1. Stiochiometry and balanced chemical


  4 Fe(s) + 3 O2(g) -------- 2 Fe2O3(s)

  2. Calculate Reactants and Products

4 mol Fe x 55.85 g Fe = 223.4g

Fe 1 mol Fe

3 mol O2 x 32.00 g O2 = 96.0g O2

1 mol O2

  Total Mass in grams of Reactants = 319.4 g


2 mol Fe2O3 x 158.7g Fe2O3= 319.4g Product

1 mol Fe2O3

Relationship derived from a balance relationship

Iron Oxygen Iron III Oxide

4 Fe(s) + 3O2(g) = 2 Fe2O3(s)

 4 atoms Fe  3 molecules O2  2 Formula Units Fe2O3

 4 moles Fe  3 moles O22 moles Fe2O3

 223.4g Fe  96.0g O2319.4g Fe2O3

319.4 g Reactants = 319.4 g Product

 Example Problem Page 355

 Practice Problems 1 (A-E) Page 356


3. Moles Ratio's

2 Al(s) + 3Br2(l) ------------ 2AlBr3(s)

2 mol Al and 2 mol Al

3 mol Br2 2 mol AlBr3

3 mol Br2 and  3 mol Br2

2 mol Al 2 mol AlBr3

2 mol AlBr3 and 2 mol AlBr3

2 mol Al 3 mol Br2

4. Practice Problems 2&3 Page 357

9.2Stiochiometric Calculations

A. Using Stiochiometry

1. Stiochiometric Moles-to-Mole Conversions

 2. Example

2K(s) + 2H2O(l) -------------- 2KOH(aq) + H2

Given= 0.0400 moles potassium Unknown= moles of Hydrogen


1 mol of H2

  2 mol K

 moles of known x moles of unknown = moles of unknown

moles of known

  0.0400 mol K x 1 mol of H2 = 0.0200 mol H2

  2 mol of K

  If you put 0.0400 mol of K into water, 0.0200 mol of H2 will be produced.

Example Problem 12-2 Page 359

 Sample Problems 9&10 Page 359

3. Stiochiometric mole-to-mass conversion

Known = moles of Chlorine = 1.25 mol Cl2

Unknown = mass of Sodium Chloride = ? g NaCl


Example Problem: Determine the mass of sodium chloride or table salt (NaCl) produced when 1.25 moles of chlorine gas reacts vigorously with sodium.

  Known = 1.25 molCl2

  Unknown = mass of Sodium chloride

 Step 1 Write a balanced equation

  2Na(s) + Cl2(g) ----------- 2NaCl(s)

 Step 2 Write the mole ratio that relates molNaCl to Cl2 2 molNaCl

1 mol Cl2

 Step 3 Multiply the number of moles of Cl2 by the mole ratio

1.25 mol Cl2 x 2 molNaCl = 2.50 molNaCl

1 mol Cl2

 Step 4 Multiply molNaCl by the molar mass of NaCl

  2.50 molNaCl x 58.44 g NaCl = 146 g NaCl

  1 molNaCl


4. Stiochiometric mass-to-mass conversion

 Example Problem: Ammonium nitrate (NH4NO3), an important fertilizer, produces N2O gas and H2O when it decomposes. Determine the mass of H2O produced from the decomposition of 25.0 g of solid ammonium nitrate.

 Known = 25.0 g NH4NO3 Unknown = mass of H20

Step 1 Write a balanced equation

 NH4NO3(s) ---------- N2O(g) + 2H2O(l)

 Step 2 Convert grams to moles

   25.0g NH4NO3 x 1mol NH4NO3 = 0.312 mol NH4NO3

80.4g NH4NO3

 Step 3 Mole Relationship

Step 4 Multiply mol NH4NO3 by the mole ratio

0.312 molNH4NO3 x 2 mol H20   = 0.624 mol H20

  1 mol NH4NO3


Step 5 Calcualte the mass of H20 using molar mass conversion factor

  0.624 mol H20 x 18.0 g H2O = 11.2 g H2O

1 mol H2O

Practice Problems 13-14 Page 362

B. Steps in Stiochiometric Calculations

1. Write a balanced chemical equation. Interpret the equation in terms of moles.

2. Determine the moles of the given substance using a mass-to- mole conversion. Use the inverse of the molar mass as the conversion factor.

3. Determine the moles of the unknown substance from the moles of the given substance. Use the appropriate mole ratio from the balanced chemical  equation as the conversion factor.

4. From the moles of the unknown substance, determine the mass of the unknown substance using a mole-to-mass conversion. Use the molar mass as the conversion  factor.


9.3 Limiting Reactants

A. Limiting reactants - limits the extent of the reaction and, thereby, determines the amount of product.

 1. A portion of all of the other reactants remains after the reaction stops.

2. Excess reactants - the left over reactants.

Before reaction

3 Nitrogen molecules  3 Hydrogen molecules   2 Amonia   2 Nitrogen Molecules

(6-Nitrogen atoms) (6-Hydrogen atoms) Molecules (4-Nitrogen atoms)

B. Calculating the product when a reactant is limited

1. Formation of disulfur dichloride (S2Cl2)

a) S8(l) + 4Cl2(aq) --------------- 4S2Cl2(l)

2. If 200-g of sulfur reacts with 100-g of chlorine, what mass of disulfurdichloride is produced?

a) You must first determine which one is the limiting reactant because the reaction will stop producing product when the limiting reactant is used up.


b) Identify the limiting reactants involves finding the number of moles of each reactant.

  100g Cl2 x 1 mol Cl2 = 1.410 mol of Cl2

70.91g Cl2

 200g S8 x 1 mol S8= 0.7791 mol S8

265.5g S8

c) Next step is to determine if the two reactants are in correct mole ratio as given in the balanced chemical equation.

  d) Equation shows a 4:1 ratio (4 moles of Cl and 1 mole of S)

  e) To determine the actual ratio, divide the available moles of chlorine by the availible moles of sulfur.

1.410 mol of Cl2 available = 1.808 mol of Cl2 available

0.7797 mol of S8 available 1 mol S8 available

  f) Only 1.808 mol of Chlorine is actually available for every 1 molof sulfur instead of the 4 moles of the chlorine required by the balanced chemicalequation. (Therefore chlorine is the limiting reactant)


12.4 Percent of Yield

A. How much product?

1. Theoretical Yield – is the maximum amount of product that can be produced from a given amount of reactant.

2. Actual Yield – is the amount of product actually produced when the chemical reaction is carried out in an experiment.

3. Percent Yield – is the ratio of the actual yield to the theoretical yield expressed as a percent.

Percent Yield = actual yield (from an experiment) x 100

theoretical yield (from stiochiometric calculations)

Example: 75 shots / made 49 = 49/75 x 100 = 65%

B. Example Problem

When potassium chromate (K2CrO4) is added to a solution containing 0.500-g of silver nitrate (Ag(NO3)2) is formed.

a) Determine the theoretical yield of the silver chromate precipitate.

b) If 0.455-g of silver chromate is obtained, calculate the percent yield.



Mass of silver nitrate = 0.500-g AgNO3

Actual yield = 0.455-g Ag2CrO4


Theoretical yield = ? Ag2CrO4

Percent yield = ? Ag2CrO4

1. Write the balanced chemical equation and indicate the known and unknown quantities.

2AgNO3(aq) + K2CrO4(aq) Ag2CrO4(s) + 2KNO3(aq)

.500-g ?

2. Convert grams of AgNO3 to moles of AgNO3, using the inverse of molar mass.

.500-g AgNO3 x 1 mol AgNO3 = 2.94 x 10-3mol AgNO3

1.69-g AgNO3


3. Use the appropriate mole ratio to convert mol AgNO3 to mol of Ag2CrO4.

2.94 x10-3mol AgNO3 x 1mol Ag2CrO4 = 1.47 x 10-3mol AgCrO4

2 mol AgNO3

4. Calculate the mass of Ag2CrO4 (theoretical yield) by multiplying mol of Ag2CrO4 by the molar mass.

1.47 x 10-3mol Ag2CrO4 x 331.7-g Ag2CrO4 = 0.488-g Ag2CrO4

1 mol Ag2CrO4

5. Divide the actual yield by the theoretical yield and multiply by 100.

0.455-g Ag2CrO4 x 100 = 93.2% Ag2CrO4

0.488-g Ag2CrO4

Practice Problems (27-29) Page 372