Stoichiometry The Mathematics of Chemical Equations.

Stoichiometry

The Mathematics of Chemical Equations

9.1: Balanced Chemical Equations

1. Provides qualitative and quantitative information.

2. Supports the Law of Conservation of MassConservation of Mass3. 2H2 + O2 → 2H2O

The above equation is interpreted in terms of particles as follows:

A. 2 molecules2 molecules of H2 react with 1 molecule1 molecule of O2 to produce 2 molecules2 molecules of water.

The ratio of H2 to O2 to H2O is 2:1:22:1:2. OR

2H2 + O2 → 2H2O

B. 20 molecules of H2 react with 1010 molecules of O2 to produce 2020 molecules of water.

Again, the ratio of H2 to O2 to H2O is 2:1:22:1:2.

C. The original equation can then be interpreted as follows:

2 moles of H2 react with 11 mole of O2 to produce 22 moles of water.

The ratio of H2 to O2 to H2O is 2:1:22:1:2.

2H2 + O2 → 2H2O

D. It is more convenient to interpret the coefficients as the number of moles, because we measure amounts of substances by gramsgrams in the laboratory. We can convert between grams and moles in solving problems involving chemical reactions.

4. Stoichiometry: The study of the mathematical (quantitative) mathematical (quantitative) relationshipsrelationships that exist in a formula and in a chemical reaction.

5. The word Stoichiometry comes from the Greek word “stoicheion” meaning elementelement, and “metron”, meaning measuremeasure.

6. Importance: The safe, economical and reproduciblereproducible manufacture of chemicals (or food) and the safe administration of pharmaceuticals.

7. Proof: The Law of conservation of mass is shown by the balancedbalanced equation, or by adding up all the masses of reactants and products to determine if they are equal.

8. Mole-Mole Problems require 3 steps to determine the molar ratio: Stoichiometryville MapA. Write Write EquationEquationB. Balance Balance the equation to determine the the equation to determine the molar ratiomolar ratio

C. Moles Given (Known) to Moles Moles Given (Known) to Moles UnknownUnknown

Volume Volume

22.4 L 22.4 L

Mass Mass

Particles Particles

Molar MassMolar Mass

6.02 x 10236.02 x 1023

Given Substance

Substance to be determined

Ex #1) Ammonium nitrate decomposes into dinitrogen monoxide and water. How many moles of products are produced from 2.25 moles of reactants?

Equation: NH4NO3 N2O + 2 H2O

Ratio: 1 : 1 : 2

2.25 mol NH4NO3 2

4 3 24 3

1 mol NO2.25 mol NH NO 2.25 mol N O

1 mol NH NO

Ex #2) Hydrochloric acid reacts with zinc in a single replacement reaction, how many moles of HCl are needed to react with 2.3 moles of Zn?

Equation:Ratio:

Stoichiometry Problems:1. A balanced equation represents a chemical

reaction that conforms to the Law of Law of Conservation of MassConservation of Mass

2. SubscriptsSubscripts can NEVER be changed to balance a chemical equation.

3. The total atoms/massatoms/mass of the reactants is the same as the total atoms/massatoms/mass of the products.

4. Coefficients in a balanced equation relate molesmoles of substances in the reaction.

5. Coefficients are used in constructing the molar molar ratioratio for a stoichiometric problem.

6. Grams of the known must first be converted to molesmoles, then use the molarmolar ratioratio to relate the 2 substances, finally convert the moles of the unknown to gramsgrams.

• B. Mass to Mass Problems1)What mass of aluminum oxide can be

prepared by the reaction of 67.5 g of aluminum and oxygen?

Al + O2 Al2O3

5. The limiting reactant is the startingstarting substance that becomes used up firstfirst when a chemical reaction occurs. It controls how muchmuch or how littlelittle product can be formed.

Ex) If a recipe for 1 cake is: and your kitchen contains

2 cups flour 7 cups flour2 eggs 2 dozen eggs1 cup sugar 9 cups sugar1½Tbls. baking powder 10 Tbls. baking

pwdr1 cup water unlimited water⅓ cup oil 3⅓ cups oil

• How many cakes can you make given the ingredients present in your kitchen? 3 What ingredient is the limiting reactant? flour How many eggs are left over after making the cakes? 18

6. 44 NH3 + 55 O2 → 44 NO + 66 H2O

33

3 3

1 mol NH 4 mol NO 30 g NO3.25 g NH = 5.74 g NO

17 g of NH 4 mol NH 1 mol NO

22

2 2

1 mol O 4 mol NO 30 g NO3.50 g O = 2.63 g NO

32 g of O 5 mol O 1 mol NO

What is the most NO you can make? ______Which reactant is the limiting reactant? ________

2.63 goxyg

en

3.25g

3.50g

? g

Reactants → Products