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3 ways of representing the reaction of H2 with O2 to form H2O

A process in which one or more substances is changed into one or more new substances is a chemical reaction.A chemical equation uses chemical symbols to show what happens during a chemical reaction: reactants products

Chemical Reactions and Chemical Equations

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How to “Read” Chemical Equations

2 Mg + O2 2 MgO

2 atoms Mg + 1 molecule O2 makes 2 formula units MgO

2 moles Mg + 1 mole O2 makes 2 moles MgO

48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO

NOT2 grams Mg + 1 gram O2 makes 2 g MgO

Chemical Reactions and Chemical Equations

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Balancing Chemical Equations1. Write the correct formula(s) for the reactants on the left

side and the correct formula(s) for the product(s) on the right side of the equation.

Ethane reacts with oxygen to form carbon dioxide and water

C2H6 + O2 CO2 + H2O

2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.

2C2H6 NOT C4H12

Chemical Reactions and Chemical Equations

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Balancing Chemical Equations3. Start by balancing those elements that appear in only one

reactant and one product.

C2H6 + O2 CO2 + H2O start with C or H but not O

2 carbonon left

1 carbonon right

multiply CO2 by 2

C2H6 + O2 2CO2 + H2O

6 hydrogenon left

2 hydrogenon right

multiply H2O by 3

C2H6 + O2 2CO2 + 3H2O

Molecules, Atoms, and Ions

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Balancing Chemical Equations

4. Balance those elements that appear in two or more reactants or products.

2 oxygenon left

4 oxygen(2x2)

C2H6 + O2 2CO2 + 3H2O

+ 3 oxygen(3x1)

multiply O2 by 72

= 7 oxygenon right

C2H6 + O2 2CO2 + 3H2O72 remove fraction

multiply both sides by 22C2H6 + 7O2 4CO2 + 6H2O

Chemical Reactions and Chemical Equations

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Balancing Chemical Equations

5. Check to make sure that you have the same number of each type of atom on both sides of the equation.

Reactants Products

4 C

12 H

14 O

4 C

12 H

14 O

2C2H6 + 7O2 4CO2 + 6H2O4 C (2 x 2) 4 C

12 H (2 x 6) 12 H (6 x 2)

14 O (7 x 2) 14 O (4 x 2 + 6)

Chemical Reactions and Chemical Equations

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Chemical Reactions and Chemical Equations

Write a balanced equation for the formation of Al2O3.

Solution

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Amounts of Reactants and Products

Stoichiometry: Mole and Mass method

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Stoichiometry: Mole and Mass method

1. Write balanced chemical equation

2. Convert quantities of known substances into moles

3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity

4. Convert moles of sought quantity into desired units

The food we eat is degraded, or broken down, in our bodies to provide energy for growth and function. A general overall equation for this very complex process represents the degradation of glucose (C6H12O6) to carbon dioxide (CO2) and water (H2O):

If 856 g of C6H12O6 is consumed by a person over a certain period, what is the mass of CO2 produced?

Stoichiometry: Mole and Mass method

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Solution

Step 1: The balanced equation is given in the problem.

Step 2: To convert grams of C6H12O6 to moles of C6H12O6, we

write

Step 3: From the mole ratio, we se

1 mol C6H12O6 equal to 6 mol CO2.

Therefore, the number of moles of CO2 formed is

Stoichiometry: Mole and Mass method

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Step 4: Finally, the number of grams of CO2 formed is given by

After some practice, we can combine the conversion steps

into one equation:

Stoichiometry: Mole and Mass method

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All alkali metals react with water to produce hydrogen gas and the corresponding alkali metal hydroxide.

A typical reaction is that between lithium and water:

How many grams of Li are needed to produce 9.89 g of H2? Lithium reacting with

water to produce

hydrogen gas.

Stoichiometry: Mole and Mass method

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From the equation we see that 2 mol Li ≡ 1 mol H2. The conversion steps are

Combining these steps into one equation, we write

Stoichiometry: Mole and Mass method

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Oxidation-Reduction Reactions

(electron transfer reactions)

2Mg 2Mg2+ + 4e-

O2 + 4e- 2O2- Reduction half-reaction (gain e-)

Oxidation half-reaction (lose e-)

2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e-

2Mg + O2 2MgO

Molecules, Atoms, and Ions

Chemical Reactions and Chemical Equations

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Molecules, Atoms, and Ions

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Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s)

Zn is oxidizedZn Zn2+ + 2e-

Zn is the reducing agent

Cu2+ is reducedCu2+ + 2e- Cu

Cu2+ is the oxidizing agent

Chemical Reactions and Chemical Equations

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Oxidation numberThe charge the atom would have in a molecule (or anionic compound) if electrons were completely transferred.

1. Free elements (uncombined state) have an oxidation number of zero.

Na, Be, K, Pb, H2, O2, P4 = 0

2. In monatomic ions, the oxidation number is equal to the charge on the ion.

Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -23. The oxidation number of oxygen is usually –2. In H2O2

and O22- it is –1.

4.4

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4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.

6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.

5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1.

7. Oxidation numbers do not have to be integers. The oxidation number of oxygen in the superoxide ion, O2

-, is –½.

Chemical Reactions and Chemical Equations

Assign oxidation numbers to all the elements in the following compounds and ion:

(a) Li2O

(b) HNO3

(c)

Chemical Reactions and Chemical Equations

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Solution

(a) By rule 2 we see that lithium has an oxidation number of +1 (Li+) and oxygen’s oxidation number is −2 (O2−).

(b) This is the formula for nitric acid, which yields a H+ ion and a N ion in solution. From rule 4 we see that H has an oxidation number of +1. Thus the other group (the nitrate ion) must have a net oxidation number of −1. Oxygen has an oxidation number of −2, and if we use x to represent the oxidation number of nitrogen, then the nitrate ion can be written as so that

x + 3(−2) = −1

x = +5

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(c) we see that the sum of the oxidation numbers in the dichromate ion must be − 2. We know that the oxidation number of O is − 2, so all that remains is to determine the oxidation number of Cr, which we call y.

The dichromate ion can be written as

so that

2(y) + 7(−2) = −2

y = +6

Chemical Reactions and Chemical Equations

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The Oxidation Numbers of Elements in their Compounds

Chemical Reactions and Chemical Equations

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Oxidation numberThe charge the atom would have in a molecule (or anionic compound) if electrons were completely transferred.

1. Free elements (uncombined state) have an oxidation number of zero.

Na, Be, K, Pb, H2, O2, P4 = 02. In monatomic ions, the oxidation number is equal to

the charge on the ion.

Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -23. The oxidation number of oxygen is usually –2. In H2O2

and O22- it is –1.

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4. The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1.

6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion.

5. Group IA metals are +1, IIA metals are +2 and fluorine is always –1.

Chemical Reactions and Chemical Equations