# Equations & Quantitative Relationships. + O 2(g)» H 2» 2 H 2

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1 Woods Chem-1 Lec Stoichiometry (std)+ Page 1 States of matter: (s) (l) (g) (aq) Balancing Equations: Equations & Quantitative Relationships H 2 (g) + O 2(g)» H 2 Can t change H 2 O to H 2 O 2 2 O (2) 1 (2) 2 H 4 2 H 2 (g) + O 2(g)» 2 H 2 Coefficients: The number of moles of each compound Subscripts: Can t change them or it will change the compound Compounds: Can t add compounds to make equation balanced (i.e. can t add O 2 as a product) Conservation of Mass: All atoms must be equal on both sides of the equation. Reaction Arrows: Favor Products Favor Reactants Reversible Polyatomic Ions: Al + H 3 PO 4» H 2 + AlPO 4 NH 3 + Cl 2» NH 4 Cl + NCl 3 C 4 H 10 + O 2» CO 2 O

2 Woods Chem-1 Lec Stoichiometry (std)+ Page 2 C 4 H 9 OH + O 2» CO 2 O C 4 H 7 OH + O 2» CO 2 O Try with Ozone (O 3 ) instead of O 2 : This is not a valid reaction, but will help to see how to clear out an x / 3 fraction! C 4 H 7 OH + O 3» CO 2 O 1) Combination Reactions A + B» AB 2) Decomposition Reactions AB» A + B

3 Woods Chem-1 Lec Stoichiometry (std)+ Page 3 3) Oxidation-Reduction Reactions Oxidation numbers are either real charges or formal charges which help chemists keep track of electron transfer. In practice, oxidation numbers are best viewed as a bookkeeping device. Oxidation cannot occur without reduction. Oxidized increase in oxidation number (LOSS of Electrons or Gain of Oxygen or Loss of Hydrogen) Reduced decrease in oxidation number (GAIN of Electrons or Loss of Oxygen or Gain of Hydrogen) Oxidizing Agent oxidizes something else: It is reduced! Reducing Agent reduces another reactant: It is oxidized. What is Oxidized, what is Reduced? Transfer of electrons: ½ Reactions: Shift of electrons: 2 Mg + O 2 2 MgO Mg Mg e O e 2 O 2 H 2 + Cl 2 2 HCl e Closer to Chlorine CH 4 + O 2 CO H 2 O e Closer to Oxygen Assigning Oxidation Numbers The oxidation number of any pure element is zero. (H in H 2 is zero) The oxidation number of a monatomic ion is equal to its charge. (Cl = 1, Mg 2+ = +2, O 2 = 2) Alkali metals are +1 Alkali earth metals are +2 Fluorine is always 1 Oxygen (O) is 2 in most compounds. Exceptions are O 2 (O = 0) and Peroxides, such as H 2 O 2 or Na 2 O 2 (O = 1), & Fluorides, such as OF 2 (O = +2) H is +1 with non-metals. Exceptions are H 2 (H = 0) and ionic Hydrides with metals, such as NaH (H = 1). The sum of the oxidation numbers is zero if neutral, or equal to the charge if an ion. For other elements, you can usually use the sum rule above to solve for the unknown oxidation number.. O 2 1 Cl as an ION is always 1 But Cl has many Oxidation Numbers O 2 Cl 7+ O 2» Cl 2(g) + O 2(g)» ClO 4 _ O 2 _ Is Cl oxidized or reduced in this example? What about O. Different oxidation #'s: Cl ClO ClO 2 NO NO 2 SO 2 4 K 2 Cr 2 O 7 NO 2 NO 3 Oxidation or Reduction? ClO 3 ClO 4

4 Woods Chem-1 Lec Stoichiometry (std)+ Page 4 Single Displacement Reactions & the Activity Series A + BC B + AC A + BC C + BA (Metals) (Halogens) Metals: Li, K, Ba, Sr, Ca, Na, Mg, Al, Mn, Zn, Cr, Fe, Co, Ni, Sn, Pb, H, Cu, Ag, Hg, Pt, Au (Most Easily Oxidized to the Left) Halogens: F 2 > Cl 2 > Br 2 > I 2 F 2 most easily reduced Greatest Electronegativity Greater pull on electrons More Easily Oxidized elements are Higher up the scale: (Don t react w/acids) Cold H 2 O H 2(g) Steam H 2 O (g) H 2(g) Acids H 2(g) No H 2(g) Very Reactive Weakly Reactive Li Mg Co H K Al Ni Cu Ba Mn Sn Ag Sr Zn Pb Hg Ca Cr Pd Na Fe Pt Au This is why Gold (Au) and Platinum (Pt) are such useful industrial compounds (eg. Gold plated wire). They are great conductors and very resistant to corrosive materials. Metals from higher up the series are most easily Oxidized, & therefore Reduce metal ions lower down the scale. Think of it as Li is at the top of an electrochemical energy Hill and Au is at the bottom. Metals from higher up the scale can pass electrons to ions of elements further Down the Hill. eg. Zn (s) has all of its electrons and could pass them Down the Hill to a Cu 2+ ion (which can gain electrons). However, Cu (s) as a metal, cannot pass electrons Up the Hill to a Zn 2+ ion. Zn (s) + Cu(NO 3 ) 2(aq) Reverse Reaction Cu (s) + Zn(NO 3 ) 2(aq) Halogens Br 2(l) + 2 NaCl (aq) Cl 2(g) + 2 NaBr (aq) Others 2 Na (s) + 2 H 2 Mg (s) + 2 HCl (aq)

5 Woods Chem-1 Lec Stoichiometry (std)+ Page 5 Switching of Cations & Anions: 4) Double Displacement (Precipitation) Reactions AgNO 3 (aq) + NaCl (aq) AgCl (s) + 3 NaNO 3 (aq) The Cation from the first reactant is paired with the Anion from the second reactant and visa versa. Note also that while both of the reactants are designated as aqueous (meaning dissolved in water), one of the products is designated as a solid. If there is no precipitation, they remain as ions No Reaction. When many salts are placed in water, the pole of H 2 O surrounds the cations, the + pole of H 2 O surrounds the anions. This process literally rips apart ionic crystals, and the ions Disassociate and disperse throughout the solution ( aq ). If a cation & anion have a greater attraction to each other then to the water molecules, they will crystallize and precipitate out of solution (i.e. form a solid). Solubility Rules Soluble Ionic Compounds Exceptions 1. + Alkali Metal (Group 1A) salts, NH 4 2. Nitrates, Perchlorates, and Chlorates 3. Acetates... Ag + 4. Chlorides, Bromides, and Iodides... Ag +, Pb 2+, Cu + 2+, and Hg 2 5. Sulfates... Sr 2+, Ba 2+, Pb 2+, Hg 2+, and Hg 2+ 2 (Ca 2+ & Ag + Slightly Sol ) 6. Fluorides... Group 2A and Pb 2+ Insoluble Ionic Compounds Exceptions 1. Hydroxides... NH + 4, Group 1A and Sr 2+ (Ba 2+ & Ca 2+ Slightly Sol ) 2. Carbonates and Phosphates... NH + 4, Group 1A 3. Sulfides... NH + 4, Group 1A (Group 2A Slightly Sol )

6 Woods Chem-1 Lec Stoichiometry (std)+ Page 6 Net Ionic Equations Molecular Equation (ME): All molecules are written in their neutral molecular form. Pb(NO 3 ) 2(aq) + 2 NaCl (aq)» PbCl 2(s) + 2 NaNO 3(aq) Helps to solve stoichiometry problems, Does not accurately portray the chemical changes which occur. Total Ionic Equation (TIE or just IE): All ions in solution are written as ions. Pb NO (aq) 3 (aq) + 2 Na+ + 2 (aq) Cl» PbCl + 2 (aq) 2(s) Na+ + 2 NO 3 Equation must be balanced in both atoms and charge. Lists ALL Ions, including the spectator ions. Weak acids/bases do not dissociate into ions. SPECTATOR IONS: Ions which are not involved in chemical change: DO NOT Change State & DO NOT Change Oxidation # Net Ionic Equation (NIE): Only write compounds involved in the reaction. Pb (aq) Cl» PbCl (aq) 2(s) Remove all spectator ions. Reduce coefficients to lowest terms. Equation must be balanced in both atoms and charge. Make sure all state symbols are included. Nickel chloride + Ammonium carbonate (ME) (IE) (aq) (NIE) Silver chlorate + Aluminum chloride (ME) (IE) (NIE) 5) Neutralization (Double Displacement) Reactions HCl (aq) + NaOH (aq)» NaCl (aq) Three Acid Rules 1) Acid + Metal - OH» Salt 2) Acid + Metal - HCO 3 or CO 3» Salt + CO 2(g) Carbonic Acid: 3) Acid + Metal» Salt (g) Acid (HCl) in stomach:

7 Woods Chem-1 Lec Stoichiometry (std)+ Page 7 Stoichiometry Dimensional Analysis: Given & Unknown. Can't go directly from one compound to another. Must go through moles. Hydrochloric acid + Calcium hydroxide Calcium chloride + Water PROB) What mass of Ca(OH) 2 (74.10 g/mol) will react with 219 g HCl (36.46 g/mol)? PROB) What mass of Ca(OH) 2 is needed to produce g CaCl 2 (111.0 g/mol)? Molarity "M" Mol / Liter [X ] mol mol Molarity: = L sol'n 1000 ml sol'n mass solute Mass-Volume %: x100 volume sol'n = g 100 ml mass solute % by Mass (m/m): x100 mass sol'n volume solute % by Volume (m/m): x100 volume sol'n Dilution Problems PROB) How would you prepare 250 ml of 2.00 M solution from a M stock solution? (Dilutions are often required in a lab, since we order concentrated solutions and then dilute out as needed). Solution Problems PROB) What mass of Ca(OH) 2 will react with 750 ml of 8.00 M HCl? PROB) How many ml of 4.00 M Ca(OH) 2 will react with 750 ml of 8.0 M HCl? % Yield Theoretical Yield: The quantities of product if a reaction is 100% complete (the stoichiometric calculation). Experimental Yield: The actual (experimental) quantity produced in an experiment. Experimental Yield % Yield = 100 % Theoretical Yield ( ) PROB) How much CaCl 2 is produced if you start with g of Ca(OH) 2 and get a 75.0 % yield? PROB) If you want 225 grams of CaCl 2 and get an 87.0% yield, how much Ca(OH) 2 is needed?

8 Woods Chem-1 Lec Stoichiometry (std)+ Page 8 Limiting Reactants Limiting Reactant: The reactant which limits (or ends) a reaction when it is used up. PROB) What is the limiting reactant and how many moles of each compound remain after the reaction is complete if you start with 6 mol HCl and 5 mol Ca(OH) 2? PROB) Use the equation: 2 A + 5 B 4 C + 3 D How much of everything would remain after reaction is complete if you started with 10 mol A & 6 mol B? PROB) If you have 222 g Ca(OH) 2 and 1.50 L of 3.50 M HCl, (1) What is the limiting reactant; (2) How much CaCl 2 (111.0 g/mol) is produced; PROB) What mass of Sodium carbonate will neutralize ml M HCl? REMEMBER: A common mistake is to try to include g/mol of HCl in the equation - IT IS NOT NEEDED PROB) A ml solution contains 60.0 g Sodium hydroxide. How much 4.0 M Sulfuric acid is needed to neutralize ml of the Sodium hydroxide? PROB) You have 39.9 g Iron (III) oxide (159.7 g/mol) and 15.0 g Carbon monoxide; (1) Which is limiting, (2) How much excess, LATER (3) How much Fe produced? Fe 2 O 3(s) + 3 CO (g) 2 Fe (l) + 3 CO 2(g) PROB) 4.65 g of Strontium hydroxide (121.6 g/mol) is added to 75.0 ml water. It takes 43.5 ml of hydrobromic acid to titrate the solution. What is the molarity of the hydrobromic acid? PROB) What mass of CO 2 would be produced if you combusted g of glucose (C 6 H 12 O 6 )? PROB) If you mix ml of 6.50 M HCl with ml of 8.20 M Ca(OH) 2 what is the limiting reactant? What mass of salt will be produced? PROB) What mass of Iron (III) oxide (159.7 g/mol) is needed to produce g Iron (III) phosphate (150.8 g/mol)? Iron (III) oxide + Phosphoric acid» Iron (III) phosphate + Water PROB) From the earlier problem with Iron (III) oxide (159.7 g/mol) and Carbon monoxide, if you needed 76.5 g of Fe and only got a 86.0% yield on the reaction, how many grams of CO would you have needed to start with?

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