III Equation Problems A What does an equation represent? Represents chemical change Made up “reactants” and “products”  Reactants  starting materials  left side  Products  end results  right side K + O 2 → K 2 O ReactantsProducts

B Balancing Equations Law of Conservation of Mass  Matter cannot be created or destroyed  Reactants equal products  Use coefficients to balance equation K + O 2 → K 2 O 1 K 2 2 O 1 4 K + O 2 → 2 K 2 O

HgO → Hg + O 2 1 Hg 1 1 O 2 2 HgO → 2 Hg + O 2

CaCl 2 + AgNO 3 → Ca(NO 3 ) 2 + AgCl 1 Ca 1 2 Cl 1 1 Ag 1 1 NO 3 2 CaCl AgNO 3 → Ca(NO 3 ) AgCl

C 2 H 4 + O 2 → CO 2 + H 2 O 2 C 1 4 H 2 2 O 3 C 2 H O 2 → 2 CO H 2 O

C 2 H 6 + O 2 → CO 2 + H 2 O 2 C 1 4 H 2 2 O 3 2 x (C 2 H O 2 → 2 CO H 2 O) 2 C 2 H O 2 → 4 CO H 2 O

C. Classification  5 Basic Types of Reactions  Synthesis  Decomposition  Single Replacement  Double Replacement  Combustion 1. Synthesis  2 substances make 1 substance  2 reactants---1 product  K + O 2 → K 2 O

2. Decomposition  1 compound makes 2 substances  1 reactant---2 products  2 HgO → 2 Hg + O 2 3. Single Replacement  1 element and 1 compound become 1 new element and 1 new compound  2 reactants---2 products  Ca + 2 AgNO 3 → Ca(NO 3 ) Ag

4. Double Replacement  2 compounds make 2 new compounds  2 reactants---2 products  CaCl AgNO 3 → Ca(NO 3 ) AgCl 5. Combustion  1 organic compound and oxygen make CO 2 and H 2 O  2 reactants---2 products  C 2 H O 2 → 2 CO H 2 O

D. Use of Equations Shows what reacts (qualitative) and how much reacts (quantitative) Ex.2 Na + Cl 2 → 2 NaCl Qualatative Na reacts with Cl 2 to make NaCl Quantitative 2 moles Na react with 1 mole Cl 2 to make 2 moles NaCl How many moles of NaCl are made using 1.4 moles Cl 2 ?  Find ratio 1:2 or 2:1  Multiply the given info by the ratio Get the given unit to cancel out 1.4 moles Cl 2 x 2 moles NaCl = 2.8 moles NaCl 1 mole Cl 2

Ex. N H 2 → 2 NH 3 How many grams of NH 3 are made using 12 g H 2 ?  Convert to moles  Multiply the given info by the ratio  Convert to grams 12g H 2 x 1 mole H 2 x 2 moles NH 3 x 17g NH 3 2g H 2 3 moles H 2 1 mole NH 3  = 68g NH 3