1. Chapter 7.2 – Chemical Equations -chemical reactions can be described two main ways 1.word equation – write the names of the products and reactants ex. methane and oxygen yield carbon dioxide and water 2.chemical equation – uses symbols to show the relationship between the reactants and products ex. CH 4 + 2O 2 → CO 2 + 2H 2 O

2. Chapter 7.2 – Chemical Equations -the number in front of a symbol is called a coefficient -coefficients tell how many of an atom or compound are involved in a chemical reaction -the arrow → means yield or get -it points from the reactants to the products -chemical equations obey the law of conservation of mass -because of this they must have equal numbers of atoms of each element on both sides of the equation

3. Chapter 7.2 – Chemical Equations 3 steps to writing a balanced chemical equation 1.write a chemical equation by substituting the correct formulas for the names of the reactants and products ex. magnesium and oxygen yield magnesium oxide Mg + O 2 → MgO

4. Chapter 7.2 – Chemical Equations 2.balance the chemical equation one element at a time -balance elements that only appear once on each side first -usually balance hydrogen and oxygen last -only change coefficients, never subscripts Mg + O 2 → MgO

5. Chapter 7.2 – Chemical Equations 3. count the atoms of each element on both sides to be sure the equation is balanced 2Mg + O 2 → 2MgO reactantsproducts Mg O

6. Chapter 7.2 – Chemical Equations

8. law of definite proportions – a compound always contains the same elements in the same proportions regardless of how the compound is created or how much compound is made -a balanced equation gives a mole ratio – - the proportion of reactants and products in a chemical equation ex. 4Al + 3O 2 → 2Al 2 O 3 -for every 4 moles of Al need 3 moles of O 2 to make 2 moles of Al 2 O 3

9. Chapter 7.2 – Chemical Equations -the balanced equation never changes no matter how much Al or O 2 we have -if I have 8 moles of Al, I need 6 moles of O 2 to react with it to give a mole ratio of 8:6, which is the same as 4:3 -it works for volumes of gases too 2H 2 O(l) → 2H 2 (g) + O 2 (g) -mole ratio of 2:2:1 -for every 2L of H 2 we made, 1L of O 2 is also made

10. Chapter 7.2 – Chemical Equations -can convert mole ratios to mass as well 4Al + 3O 2 → 2Al 2 O 3 -Al molar mass 26.98 g/mol x 4 = 107.92 g Al -O 2 molar mass 32.00 g/mol x 3 = 96.00 g O 2 -for Al to react we need 96 g of O 2 for every 107.92 g of Al

11. Chapter 7.2 – Chemical Equations