1. Chemical Equations Honors Chemistry Unit 7

2. Writing and Balancing Equations  Chemical Reactions A.Reactants – Starting materials B.Products – Materials formed during reaction (ending materials) C.Indicate the physical state with a letter: (s) – solid, (l) – liquid, (g) – gas, (aq) – aqueous solution (dissolved in water)  A balanced equation shows conservation of mass because atoms are conserved (same on both sides).

3. When Balancing Equations:  Use coefficients in front of the formulas to balance- NEVER change subscripts once the formula has been correctly written.  Leave the “lumps” (polyatomic ions) as “lumps”.  Start with an element that only appears in one species on each side of the equation.  The simplest whole number ratio should be used.  Steps to balance: 1.Determine formulas of reactants and products. 2.Assemble the equation parts a.Reactants on left with plus sign between b.Products on right with plus sign between c.Connect with an arrow  d.Use correct formulas (review polyatomic ions) 3.Balance by changing COEFFICIENTS ONLY!! (number in front) NEVER CHANGE SUBSCRIPTS after formula has been correctly written. 4.Do an atom count to make sure they are equal on both sides.

4. Examples  Hydrogen and oxygen combine to make water. 1.Both hydrogen and oxygen are in BrINClHOF, so they are diatomic. The correct formulas are: H 2 + O 2  H 2 O 2.Indicate physical state of reactants and products: H 2(g) + O 2(g)  H 2 O (l) 3.Balance with coefficients to give same atom numbers on both sides. 2H 2(g) + O 2(g)  2H 2 O (l)  Liquid carbon disulfide reacts with oxygen to produce carbon dioxide and sulfur dioxide gases. 1. CS 2 + O 2 → CO 2 + SO 2 2.Indicate the physical states: CS 2(l) + O 2(g) → CO 2(g) + SO 2(g) 3.Balance by putting coefficients in front to give same # of each type of atom on both sides. CS 2(l) + 3O 2(g) → CO 2(g) + 2SO 2(g)  Iron (III) nitrate solution reacts with aqueous sodium hydroxide to form solid iron (III) hydroxide and aqueous sodium nitrate. (on board) Final balanced equation: Fe(NO 3 ) 3(aq) + 3NaOH (aq)  Fe(OH) 3(s) + 3NaNO 3(aq)

5. Five Basic Reaction Types  Single Displacement : one element replaces another If the single element forms cations then reaction is A + BX → AX + B (metals) or if the element forms anions (nonmetals) then X + AY → AX + Y. (also called single replacement)Will happen if single element is more reactive than the one in the compound. Less reactive is alone. 1.element + compound  element + compound 2.Examples: a.Zn + 2HCl  ZnCl 2 + H 2 (Zn is a metal : + ion joins with -) b.Cl 2 + 2HI  I 2 + 2HCl (Cl is a nonmetal: - joins with +)

6. Reaction Types (continued)  Double Displacement : Two compounds - Ions swap partners. AX + BY → AY + BX (also called double replacement) 1.compound + compound  compound + compound 2.Examples: a.Na 2 SO 4 + Ba(NO 3 ) 2  2NaNO 3 + BaSO 4 b.ZnBr 2 + 2AgNO 3 Zn(NO 3 ) 2 + 2AgBr  Decomposition : breakup of compound, requires energy AX → A + X begins with a single reactant 1.compound  two or more elements or compounds 2.Examples: a.2KNO 3  2KNO 2 + O 2 b.NH 4 NO 3  N 2 O + 2H 2 O

7. Reaction Types (continued)  Synthesis : 2 or more substances combine to form one compound A + X → AX (creates a single product) 1.Two or more elements or compounds  compound 2.Examples: a.Zn + I 2  ZnI 2 b.NH 3 + HCl  NH 4 Cl  Combustion : burning of carbon compounds (reacts with O 2 in air) 1.Hydrocarbon compound + O 2 → CO 2 + H 2 O 2.Examples: a.CH 4 + 2O 2  CO 2 + 2H 2 O (natural gas) b.2C 2 H 6 + 7O 2  4CO 2 + 6H 2 O

8. Mass Relations from Equations  Information form Chemical Equations A.The coefficients in this equation represent numbers of atoms or molecules of substances in the reaction B.Since a mole is always the same number of atoms or molecules, the coefficients also represent the numbers of moles of the substances. C.2H 2(g) + O 2(g)  2H 2 O (l)  Therefore, we can obtain conversion factors from a balanced equation. From the equation above, we know the following: 2 mol H 2 O = 2 mol H 2 = 1 mol O 2  These ratios can help us solve quantitative problems dealing with amounts of materials in a chemical reaction. (stoichiometry problems)

9. Limiting Reactant and Theoretical Yield  When a reaction goes to completion and there is no loss of product, then the amount is called the theoretical yield. This is usually determined from a stoichiometry setup.  Most of the time reactants are not present in stoichiometric amounts. We usually use an excess of the cheaper reactants. The limiting reactant is the one that runs out first.  To determine the limiting reactant: 1.Calculate the amount of products (moles or grams) that would be formed if each reactant were used up completely. This requires a separate stoichiometry setup for each reactant. 2.The theoretical yield of product is the smallest of the amounts calculated. 3.The limiting reactant is the reactant that gave this smallest amount of product.  Examples: See Board

10. Experimental Yield and Percent Yield  Some of the limiting reagent may be consumed in competing reactions; some of the product may be lost in separating it from the reaction mixture.  The amount actually produced is expressed as a percentage of the theoretical, or calculated, yield.  Percent yield = experimental yield x 100 theoretical yield  Examples on Board