3 ways of representing the reaction of H 2 with O 2 to form H 2 O

1. reactants products Chemical Reactions and Chemical Equations 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: 3 ways of representing the reaction of H2 with O2 to form H2O

2. How to “Read” Chemical Equations Chemical Reactions and 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 NOT 2 grams Mg + 1 gram O2 makes 2 g MgO

3. C2H6 + O2 CO2 + H2O NOT 2C2H6 C4H12 Chemical Reactions and Chemical Equations Balancing Chemical Equations • 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 • 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.

4. 1 carbon on right 6 hydrogen on left 2 hydrogen on right 2 carbon on left C2H6 + O2 C2H6 + O2 C2H6 + O2 CO2 + H2O 2CO2 + H2O 2CO2 + 3H2O Molecules, Atoms, and Ions Balancing Chemical Equations • Start by balancing those elements that appear in only one reactant and one product. start with C or H but not O multiply CO2 by 2 multiply H2O by 3

5. multiply O2 by 4 oxygen (2x2) + 3 oxygen (3x1) 2 oxygen on left C2H6 + O2 2CO2 + 3H2O C2H6 + O2 2CO2 + 3H2O 2C2H6 + 7O2 4CO2 + 6H2O 7 7 2 2 Chemical Reactions and Chemical Equations Balancing Chemical Equations • Balance those elements that appear in two or more reactants or products. = 7 oxygen on right remove fraction multiply both sides by 2

6. 4 C (2 x 2) 4 C Reactants Products 12 H (2 x 6) 12 H (6 x 2) 14 O (7 x 2) 14 O (4 x 2 + 6) 4 C 4 C 12 H 12 H 14 O 14 O 2C2H6 + 7O2 4CO2 + 6H2O Chemical Reactions and Chemical Equations Balancing Chemical Equations • Check to make sure that you have the same number of each type of atom on both sides of the equation.

7. Chemical Reactions and Chemical Equations Write a balanced equation for the formation of Al2O3. Solution

8. Stoichiometry: Mole and Mass method Amounts of Reactants and Products

9. Stoichiometry: Mole and Mass method • Write balanced chemical equation • Convert quantities of known substances into moles • Use coefficients in balanced equation to calculate the number of moles of the sought quantity • Convert moles of sought quantity into desired units

10. 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

11. 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

12. 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

13. 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? Stoichiometry: Mole and Mass method Lithium reacting with water to produce hydrogen gas.

14. Stoichiometry: Mole and Mass method From the equation we see that 2 mol Li ≡ 1 mol H2. The conversion steps are Combining these steps into one equation, we write

15. 2Mg 2Mg2+ + 4e- O2 + 4e- 2O2- 2Mg + O2 2MgO 2Mg + O2 + 4e- 2Mg2+ + 2O2- + 4e- Oxidation-Reduction Reactions Molecules, Atoms, and Ions (electron transfer reactions) Oxidation half-reaction (lose e-) Reduction half-reaction (gain e-)

16. Chemical Reactions and Chemical Equations

17. Molecules, Atoms, and Ions

18. Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s) Cu2+ + 2e- Cu Zn Zn2+ + 2e- Chemical Reactions and Chemical Equations Zn is oxidized Zn is the reducing agent Cu2+is reduced Cu2+ is the oxidizing agent

19. Oxidation number Chemical Reactions and Chemical Equations The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. • Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4 = 0 • In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 • The oxidation number of oxygen isusually–2. In H2O2 and O22- it is –1. 4.4

20. Chemical Reactions and Chemical Equations • The oxidation number of hydrogen is +1except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. • Group IA metals are +1, IIA metals are +2 and fluorine is always –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. • Oxidation numbers do not have to be integers. The oxidation number of oxygen in the superoxide ion, • O2-, is –½.

21. Assign oxidation numbers to all the elements in the following compounds and ion: Li2O HNO3 Chemical Reactions and Chemical Equations

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

23. (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

24. Chemical Reactions and Chemical Equations The Oxidation Numbers of Elements in their Compounds

25. Chemical Reactions and Chemical Equations Oxidation number The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. • Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H2, O2, P4 = 0 • In monatomic ions, the oxidation number is equal to the charge on the ion. Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2 • The oxidation number of oxygen is usually–2. In H2O2 and O22- it is –1.

26. Chemical Reactions and Chemical Equations • The oxidation number of hydrogen is +1except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. • Group IA metals are +1, IIA metals are +2 and fluorine is always –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.