Chemical Reactions



Unit VI – Stoichiometry and Redox Notes

Stoichiometry

• Chemical reactions will stop when one of the reactants is used up (unless it is a reversible reaction)

o Stoichiometry is the tool to measuring the amount of reactants needed to have them all used up in the end

o The study of quantitative relationships between amounts of reactants used and products formed by chemical reactions

• It is based on the Law of Conservation of Mass, which was introduced by Antoine Lavoisier in the 18th century

o Recall: Law of Conservation of Matter/Energy

▪ Mass is neither created nor destroyed in a chemical reaction, but preserved throughout

▪ Chemical bonds in reactants are broken and new chemical bonds form to produce products, but the amount of matter/mass/energy present at the end of the reaction is the same as was present at the beginning

Determining the Mass of All Reactants:

4Fe + 3O2 ( 2Fe2O3

(4 moles of Fe x 55.85g) = 223.4 g Fe

(3 moles of O2 x 32 g) = 96.00 g O2

So 223.4g Fe + 96.00 g O2 = 319.4 g or the reactants are 319.4 g

Determining the Mass of All Products:

4Fe + 3O2 ( 2Fe2O3

(2 moles of Fe2O3 x 159.7 g Fe2O3 = 319.4 g

The product = 319.4 g so the reactants = 319.4 g = products

Mole Ratios

• Coefficients in chemical reactions also indicate the relationships between reactants and products

o In other words… 4Fe + 3O2 ( 2Fe2O3

▪ 4 mol Fe will react with 3 mol oxygen gas to produce 2 mol IronIII oxide

o Mole Ratio: The ratio between the number of moles of any two substances in a BALANCED chemical equation

Example:

4Fe + 3O2 ( 2Fe2O3

o 4 mol Fe / 3 mol O2

o 4 mol Fe / 2 mol Fe2O3

o 3 mol O2 / 2 mol Fe2O3

• You would do this to determine a certain amount of product gained from a reactant

o In other words… how much reactant do I use?

Example:

Mole to Mole Stoichiometry:

2 K + 2H2O 2KOH + H2

• Figure out how many moles of hydrogen gas is produced with 0.0400 mol K

What you know:

Given 0.0400 mol K, 2 mol K = 1 mol H2

So… 0.0400 mol K * 1 mol H2O = 0.02 mol H2

1 2 mol K

Another Example:

C3H8 + 5O2 ( 3CO2 + 4H2O

• How many moles of carbon dioxide and water will be made from 10 mol propane (C3H8)?

10 mol C3H8 * 3 mol CO2 = 30 mol CO2

1 1 mol C3H8

10 mol C3H8 * 4 mol H2O = 40 mol H2O

1 1 mol C3H8

Mole to Mass Stoichiometry

2 Na + Cl2 ( 2NaCl

• Determine the mass of salt produced with 1.25 mol chlorine gas

1.25 mol Cl2 * 2 mol NaCl * 58.44g NaCl

1 1 mol Cl2 1 mol NaCl

As always, multiply everything on top and divide by everything on the bottom…

146 g NaCl from 1.25 moles of chlorine gas

• Now calculate how many grams of salt will be produced if I gave you 25 g of sodium

2 Na + Cl2 ( 2NaCl

25 g Na * 1 mol Na * 2 mol NaCl * 58.44g NaCl =

1 22.99 g Na 2 mol Na 1 mol NaCl

(25 * 1 * 2 * 58.44) / (1 * 22.99 * 2 * 1) = 64 g NaCl

Step by Step Procedure

1. Start with a BALANCED EQUATION

a. Interpret the equation in terms of moles

2. Convert from grams to moles

a. Use the inverse of the molar mass as the conversion factor

3. Convert from moles of given to moles of unknown substance

a. Uses the appropriate mole ratio from the balanced equation as the conversion factor

4. Convert from moles to grams

a. Use the molar mass as the conversion factor

Limiting Reactants - The reason why chemical reactions stop

• Limiting reactant: substance that limits the extent of the reaction and thereby determines the amount of products

• Excess Reactants: left-over reactants

These will be calculated using stoichiometry

Example:

S8 + 4Cl2 4S2Cl2

• If 200 g of sulfur reacts with 100 g of chlorine, what mass of disulfur dichloride is produced?

100 g Cl2 * 1 mol Cl2 * 4 mol S2Cl2 * 135 g S2Cl2 =

1 70.91 g Cl2 4 mol Cl2 1 mol S2Cl2

This equals 190.4 g S2Cl2

200 g S8 * 1 mol S8 * 4 mol S2Cl2 * 135 g S2Cl2 =

1 256.5 g S8 1 mol S8 1 mol S2Cl2

This equals 421.1 g S2Cl2

Since S8 will make 421.1 g of S2Cl2 and Cl2 will only make 190.4 g, therefore Cl2 will be the limiting reactant because it is used up before S8 can make 421.1 g

Why Use an Excess Reactant

1. Some reactions will actually stop with low concentrations of reactants so an excess is used to completely use the limiting reactant

o When this is done in the laboratory, use the least expensive reactant for the excess reactant

2. It can speed up a chemical reaction

o (Remember concentrations!)

Percent Yield

• Now since you are able to calculate the amount of product that should be obtained from the reaction…

• You must be able to show the percent that is physically obtained from the experiment

o In other words, if you used stoichiometry and calculated 20g of product with the limiting reactant, but only got 19g. What is your percent yield?

Percent Yield = (actual yield (from experiment)) * 100%

(Theoretical Yield (stoichiometric calculations))

Redox Reactions

• One of the defining characteristics of single-replacement and combustion reactions is that they always involve the transfer of electrons from one atom to another

Complete Chemical Equation:

• 2 Na(s) + Cl2(g) ( 2 NaCl(s)

Net Ionic Equation:

• 2 Na(s) + Cl2(g) ( 2 Na+ + 2 Cl-(ions in crystal)

• In this reaction, an electron is transferred from each of the two sodium atoms to the chlorine atoms to form two Cl- ions

• A reaction in which electrons are transferred from one atom to another is called an oxidation-reduction reaction

• We use Redox for simplicity

o How do oxidation and reduction differ?

o Oxidation is referred to as the loss of electrons from atoms of a substance

o Na ( Na+ + e-

o Reduction is defined as the gain of electrons by atoms of a substance

o Cl2 + 2e- ( 2Cl-

o An easy way to remember the difference is the saying:

o LEO the lion says GER or LEO GER

o Loss of Electrons is Oxidation

o Gain of Electrons is Reduction

• Remember- The oxidation number of an atom is the charge made by ions forming

• Group 1 is +1, Group 2 is +2, Group 17 is -1, etc.

• In Redox Reactions, one substance will acting as the oxidizing agent.

• Thus the substance that oxidizes another substance by accepting its electrons is called an oxidizing agent

• The reducing agent is the substance that reduces another substance by losing electrons

• A disproportionation is a substance that acts a reducing and oxidizing agent at the same time

• Ionic bonds will form ions and thus have oxidation charges

• Covalent do not have formal oxidations because they share electrons

• Thus, their oxidation numbers in redox reactions are based purely on electronegativity values in relation to the bond

Rules for Determining Oxidation Numbers

1. The oxidation number of an uncombined atom is zero. This is also true for polyatomic molecules such as O2, Cl2, S8, etc.

2. The oxidation number of a monatomic ion is equal to the charge on the ion Na1+ is +1, Cl1- is -1

3. The oxidation number of the more electronegative atom in a molecule or a complex ion is the same as the charge it would have if it were an ion

4. The most electronegative element, fluorine, always has an oxidation number of -1 when it is bonded to another element

5. The oxidation number of oxygen in compounds is always -2, except in peroxides (H2O2) and when bonded to the only element more electronegative than itself, F (it will be +2 in this case)

6. The oxidation number of hydrogen in most cases is +1, unless its bonded to a less electronegative atom such as metals in group 1, 2 and 13

7. The metals of groups 1 and 2 and Aluminum in 13 are still always +1, +2 and +3, respectively

8. The sum of the oxidation numbers in a neutral compound must be zero

9. The sum of oxidation numbers of the atoms in a polyatomic ion is equal to the charge on the ion

Balancing Redox Equations

• Simple redox reactions require the inspection technique

o The skill we have been using this year, which is done by just counting up the atoms and balancing them on both sides

o More complex redox reactions will require a more systematic (mathematical) approach

▪ Oxidation Number Method

▪ Equation-Balance Method

Oxidation Number Method

1. Assign oxidation numbers to all atoms in the equation

2. Identify the atoms that are oxidized and the atoms that are reduced

3. Determine the charge in oxidation number for the atoms that are oxidized and for the atoms that are reduced

4. Make the change in oxidation numbers equal in magnitude by adjusting the coefficients in the equation

5. If necessary, use the conventional method to balance the remainder of the equation

Ex. Cu + HNO3 → Cu(NO3)2 + NO + H2O

1. Assign oxidation numbers to all atoms in the equation

Oxidation: 0 +1 +5 -2 +2 +5 -2 +2 -2 +1 -2

Cu + H N O3 → Cu(N O3)2 + N O + H2O

2. Identify the atoms that are oxidized and the atoms that are reduced

Oxidized: Cu0 ( Cu+2 + 2e-

Reduced: N+5 + 3e- ( N+2

3. Determine the change in oxidation number for the atoms that are oxidized and for the atoms that are reduced

Copper: Net Change = +2 Nitrogen: Net Change = -3

4. Make the change in oxidation numbers equal in magnitude by adjusting the coefficients in the equation with the elements whose oxidation is changing

3Cu + HNO3 → Cu(NO3)2 + 2NO + H2O

5. If necessary, use the conventional method to balance the remainder of the equation

3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

Practice Problems:

1. Al(s)  +  MnO2(s)  -->  Al2O3(s)  +  Mn(s)

2. SO2(g)  +  HNO2(aq)  -->  H2SO4(aq)  +  NO(g)

3. HNO3(aq)  +  H2S(aq)  -->  NO(g)  + S(s)  +  H2O(l)

4. Al(s)  +  H2SO4(aq)  -->  Al2(SO4)3(aq)  +  H2(g)

Equation-Balance Method

• This method can be broken down into two subsets:

o Acidic Condition

▪ HNO3(aq)  +  H2S(aq)  --> 

▪ An acid as one of the reactants

o Basic Condition

▪ Cr(OH)3(s) + ClO3-(aq)  --> 

▪ A base as one of the reactants

Acidic Method

Step 1: Write the skeletons of the oxidation and reduction half-reactions. (They contain the formulas of the compounds oxidized and reduced, but the atoms and electrons have not yet been balanced.)

Step 2: Balance all elements other than H and O.

Step 3: Balance the oxygen atoms by adding H2O molecules where needed.

Step 4: Balance the hydrogen atoms by adding H+ ions where needed.

Step 5: Balance the charge by adding electrons, e-.

Step 6: If the number of electrons lost in the oxidation half-reaction is not equal to the number of electrons gained in the reduction half-reaction, multiply one or both of the half- reactions by a number that will make the number of electrons gained equal to the number of electrons lost.

Step 7: Add the 2 half-reactions as if they were mathematical equations. The electrons will always cancel. If the same formulas are found on opposite sides of the half-reactions, you can cancel them. If the same formulas are found on the same side of both half-reactions, combine them.

Step 8: Check to make sure that the atoms and the charges balance.

Practice Problems

1. MnO4-(aq) + Br-(aq)  -->  MnO2(s) + BrO3-(aq)

2. Mn(s) + HNO3(aq)  -->  Mn2+(aq) + NO2(g)

Basic Method

Steps 1-7: Begin by balancing the equation as if it were in acid solution. If you have H+ ions in your equation at the end of these steps, proceed to Step #8. Otherwise, skip to Step #11.

Step 8: Add enough OH- ions to each side to cancel the H+ ions. (Be sure to add the OH- ions to both sides to keep the charge and atoms balanced.)

Step 9: Combine the H+ ions and OH- ions that are on the same side of the equation to form water.

Step 10: Cancel or combine the H2O molecules.

Step 11: Check to make sure that the atoms and the charge balance. If they do balance, you are done. If they do not balance, re-check your work in Steps 1-10.

Practice Problems

1. H2O2(aq) + ClO4-(aq)  -->  O2(g) + ClO2-(aq)

2. S2-(aq) + I2(s)  -->  SO42-(aq) + I-(aq)

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