Oh, That Curious Ira Remsen!

Oh, That Curious Ira Remsen!

This demonstration has its roots in the writings of Ira Remsen, a 19th century professor of chemistry at Johns Hopkins University. In his memoir, Remsen writes.... "While reading a textbook of chemistry I came upon the statement, nitric acid acts upon copper. I was getting tired of reading such absurd stuff and I was determined to see what this meant. Copper was a more or less familiar to me, for copper cents were then in use.

"I had seen a bottle marked nitric acid on a table in the doctor's office where I was then doing time. I did not know its peculiarities, but the spirit of adventure was upon me. Having nitric acid and copper, I had only to learn what the words acts upon meant . The statement nitric acid acts upon copper would be something more than mere words. All was still. In the interest of knowledge I was even willing to sacrifice one of the few copper cents then in my possession.

"I put one of them on the table, opened the bottle marked nitric acid, poured some of the liquid on the copper and prepared to make an observation. But what was this wonderful thing which I beheld? The cent was already changed and it was no small change either. A green-blue liquid foamed and fumed over the cent and over the table. The air in the neighborhood of the performance became colored dark red. A great colored cloud arose. This was disagreeable and suffocating.

"How should I stop this? I tried to get rid of the objectionable mess by picking it up and throwing it out of the window. I learned another fact. Nitric acid not only acts upon copper, but it acts upon fingers. The pain led to another unpremeditated experiment. I drew my fingers across my trousers and another fact was discovered. Nitric acid acts upon trousers.

" Taking everything into consideration, this was the most impressive experiment and relatively probably the most costly experiment I have ever performed... It was a revelation to me. It resulted in a desire on my part to learn more about that remarkable kind of action. Plainly, the only way to learn more about it was to see its results, to experiment, to work in a laboratory."

Demonstration Demonstration
Ira Remsen Demo Cu (s) + HNO3 (aq)  CuNO3 (aq) + H2 (g) Set-up Observations Explanation Copper Penny placed in Nitric Acid Shows Evidence of a chemical reaction

Demonstration Demonstration Yellow Precipitate
2KI(aq) + Pb(NO3)2 (aq)  2KNO3 (aq) + PbI2 (s) Double Displacement Reactions

Al(s) + CuCl2 (aq)  Cu(s) + AlCl3 (aq) Single Displacement Reactions
Demonstration Demonstration Can Ripper Al(s) + CuCl2 (aq)  Cu(s) + AlCl3 (aq) Single Displacement Reactions

Evidence of a chemical reaction; energy absorbed
Cool Reaction Set-up Observations Explanation Evidence of a chemical reaction; energy absorbed

Evidence of a chemical reaction; production of light and heat
Combustion of Methane Set-up Observations Explanation Evidence of a chemical reaction; production of light and heat

Demonstration Demonstration EGGsplosive 2H2 (g) + O2(g)  2H2O(g)
Synthesis Reaction Combustion Reaction Exothermic Reactions

Demonstration Demonstration Elephant Toothpaste Decomposition Reaction
2H2O2(g) -> 2H2O(l) + O2(g) Decomposition Reaction

Methane Mamba CH4 (g) + O2 (g)  CO2 (g) + H2O(g)
Demonstration Demonstration Methane Mamba CH4 (g) + O2 (g)  CO2 (g) + H2O(g) Combustion Reactions

Chemistry Chapter 8 Chemical Equations and Reactions

Model #1 Balancing Chemical Equations
With your group read the play; each member of the group should take one of the parts. Use the information in the play to answer Key Questions

Model #1 Balancing Chemical Equations Class Check-in
1. The reactants are Propane (C3H8) and oxygen (O2) 2. The products are Carbon Dioxide (CO2) and Water (H2O) 3. The arrow means produces or yields 4.The subscripts tell how many of each atom are present in the molecular formula

Model #1 Balancing Chemical Equations Class Check-in
5. The number of molecules 6. The second balanced equations includes the states of matter for each reactant and product 7. a. The stoichiometric coefficients specify the number of molecules of each reactant and product b. The stoichiometric coefficients specify the number of moles of each reactant and product

Balancing Chemical Equations
Section 2 Balancing Chemical Equations

Balancing Chemical Equations The Game!
Model #2 Balancing Chemical Equations The Game! Goals of the Game To get the same number of atoms of each element in the reactant and the product. 2. To obey the law of conservation of mass.

Model #2 Balancing Chemical Equations The Game! Rules of the Game 1. The equation must represent known facts 2. The equation must contain the correct formula for the reactants and products Remember Diatomic Elements I2 Br2 Cl2 F2 O2 N2 H2 Remember the Molecular Elements S8 P4

Model #2 Balancing Chemical Equations The Game! Rules of the Game 3. Only coefficients can be added or changed 4. Once formulas are written subscripts can not be changed 5. Remember the states of matter s = solid aq = aqueous (dissolved in water L = liquid g = gas

Model #2 Balancing Chemical Equations The Game! Rules of the Game 6. Memorize the name of common acids Hydrochloric acid HCl Nitric Acid HNO3 Hydroiodic Acid HI Sulfuric Acid H2SO4 Hydrofluoric Acid HF Phosphoric Acid H3PO4 Acetic Acid HC2H3O2

Model #2 Balancing Chemical Equations The Game! Tips for Play 1. Balance the different types of atoms one at a time 2. First balance element that appear only once on each side. 3. Balance the polyatomic ions that appear on both sides as a single unit . 4. Balance Hydrogen and oxygen last

Model #2 Balancing Chemical Equations The Game! Tips for Play 5. Keep a tally for each element on each side below the equation 6. If it could be balanced by a coefficient of 1 ½ - use it – then multiply all coefficients in the equation by 2.

Balancing Chemical Equations
Model #3 Balancing Chemical Equations Solid aluminum carbide reacts with water to produce methane gas and solid aluminum hydroxide.

Practice Problem #1 Class Check-In
Step #1 aluminum carbide + water  methane + aluminum hydroxide Al4C H2O  CH Al(OH)3 12 3 4 Step #2 Al = C = H = O = 4 3 24 12 4 3 2 1 Al = C = H = O = 4 3 24 12 1 7 3 4 1 16 12 Step #3 Al4C3 (s) + 12H2O (l)  3CH4 (g) + 4Al(OH)3 (s)

Practice Problems Use Model #1- #3 to complete Practice Problems 1 – 3; write balanced chemical equations for each of the three reactions.

Write a word equation for baking a cake from scratch.
Bellringer #3 01/20/15 Write a word equation for baking a cake from scratch. Does a cake have the same properties as the ingredients?

magnesium + hydrochloric acid  magnesium chloride + hydrogen Mg HCl  MgCl H2 2 Mg = H = Cl = 1 1 2 Mg = H = Cl = 1 2 State of matter symbols State of matter symbols Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g)

nitric acid + magnesium hydroxide  magnesium nitrate + water HNO Mg(OH)2  Mg(NO3)2 + HOH 2 2 2 1 H = NO3 = Mg = OH = 1 2 H = NO3 = Mg = OH = 1 2 2 2HNO3(aq) + Mg(OH)2(s)  Mg(NO3)2(s) + 2HOH(l)

Ethane + oxygen  carbon dioxide + water C2H O2  CO2 + H2 O 2 3 3½ C = H = O = 2 6 7 2 6 C = H = O = 2 5 1 2 3 2 6 7 Multiply all coefficients by 2! 2C2H6 (g) O2 (g)  4CO2 (g) + 6H2 O (g)

Activity Series A list of elements organized according to the ease with which elements undergo certain chemical reactions Replace an element below it in the activity series Only for single displacement reactions

Activity Series Table 4 pg 281 Examples
Potassium, Aluminum, Iron, Gold 2. Fluorine, Chlorine, Bromine, Iodine

Memorize the name of common acids
Model #2 Memorize the name of common acids Hydrochloric acid HCl Nitric Acid HNO3 Hydrobromicbromic HBr Sulfuric Acid H2SO4 Hydroiodic Acid HI Phosphoric Acid H3PO4 Hydrofluoric Acid HF Acetic Acid HC2H3O2

Problems A & B

Writing and Balancing Chemical Equations
Aqueous aluminum hydroxide yields aluminum oxide and water Chemical Equation Al3+ OH- Al3+ O2- Al2O3(s) + H2O (l) Al(OH)3(aq) 

#1B Al(OH)3(aq)  Al2O3(s) + H2O(l) 2 3 Al = O = H = 2 6 1 3 Al = O =
aluminum hydroxide  aluminum oxide + water Al(OH)3(aq)  Al2O3(s) + H2O(l) 2 3 Al = O = H = 2 6 1 3 Al = O = H = 2 4 2 6 2Al(OH)3(aq)  Al2O3(s) + 3H2O(l)

Aqueous calcium fluoride reacts with aqueous hydrogen sulfate to give aqueous hydrogen fluoride and solid calcium sulfate. Chemical Equation SO42- Ca2+ F- H+ F- Ca2+ H+ SO42-  HF(aq) + CaF2(aq) + H2SO4(aq) CaSO4(s)

#2B CaF2(aq) + H2SO4(aq)  HF(aq) + CaSO4(s) 2 Ca = F = H = SO4 = 1 2
calcium fluoride + hudrogen sulfate  hydrogen fluoride + calcium sulfate CaF2(aq) + H2SO4(aq)  HF(aq) + CaSO4(s) 2 Ca = F = H = SO4 = 1 2 Ca = F = H = SO4 = 1 1 2 CaF2(aq) + H2SO4(aq)  2HF(aq) + CaSO4(s)

Solid iron (III) oxide reacts with hydrogen gas to produce iron and water Chemical Equation Diatomic! Fe3+ O2- + H2O (l) Fe2O3(s) + H2(g)  Fe(s)

#3B Fe2O3(s) + H2(g)  Fe(s) + H2O(l) 2 3 3 Fe= O = H = 2 3 Fe = O =
iron (III) oxide + hydrogen  iron + water Fe2O3(s) + H2(g)  Fe(s) + H2O(l) 2 3 3 Fe= O = H = 2 3 Fe = O = H = 1 2 2 3 6 2 3 6 2 1 Fe2O3(s) H2(g)  2Fe(s) H2O(l)

Aluminum plus aqueous manganese (II) oxide yield manganese and aqueous aluminum oxide. Chemical Equation Mn2+ O2- Al3+ O2- Mn(s) MnO(aq)  + Al2O3 (aq) Al(s) +

#4B Al(s) + MnO(aq)  Mn(s) + Al2O3(aq) 2 3 3 Al= Mn = O = 2 1 1 Al =
aluminum + manganese (II) oxide  manganese + aluminum oxide Al(s) + MnO(aq)  Mn(s) + Al2O3(aq) 2 3 3 Al= Mn = O = 2 1 1 Al = Mn = O = 2 1 3 2 3 2 3 2Al(s) + 3MnO(aq)  3Mn(s) + Al2O3(aq)

A solution of dihydrogen sulfide reacts with oxygen gas to produce water and aqueous sulfur dioxide. Chemical Equation H2S(aq) + SO2(aq) + O2(g)  H2O (l) Diatomic!

#5B H2S(aq) + O2(g)  H2O(l) + SO2(aq) 2 1 ½ 3 2 2 H= S = O = 2 1 H =
dihydrogen sulfide + oxygen  water + sulfur dioxide H2S(aq) O2(g)  H2O(l) + SO2(aq) 2 1 ½ 3 2 2 H= S = O = 2 1 H = S = O = 2 1 3 4 2 6 2 1 3 4 2 6 Multiply all coefficients by 2! 2H2S(aq) O2(g)  2H2O(l) + 2SO2(aq)

Solutions of beryllium iodide and tin(IV) nitrate yields aqueous beryllium nitrate and solid tin(IV) iodide Chemical Equation I- Be2+ Sn4+ NO3- I- Sn4+ Be2+ NO3- BeI2(aq) Sn(NO3)4 (aq) +  Be(NO3)2(aq) + SnI4(s)

#6B BeI2(aq) + Sn(NO3)4(aq)  Be(NO3)2(aq) + SnI4(s) 2 2 Be= I = Sn =
beryllium iodide + tin(IV) nitrate  beryllium nitrate + tin(IV) iodide BeI2(aq) + Sn(NO3)4(aq)  Be(NO3)2(aq) + SnI4(s) 2 2 Be= I = Sn = NO3 = 2 4 1 1 2 4 Be = I = Sn = NO3 = 1 4 2 2 4 1 2BeI2(aq) + Sn(NO3)4(aq)  2 Be(NO3)2(aq) + SnI4(s)

Solid iron reacts with fluorine gas to give solid iron (III) fluoride. Chemical Equation Diatomic Fe3+ F- Fe(s) F2(g) +  FeF3 (s)

#7B Fe(s) + F2(g)  FeF3(s) 2 3 1 ½ 2 Fe= F = 1 2 Fe = F= 1 3 2 6 2 6
Iron + Fluorine  Iron (III) fluoride Fe(s) F2(g)  FeF3(s) 2 3 1 ½ 2 Fe= F = 1 2 Fe = F= 1 3 2 6 2 6 1 3 Multiply all coefficients by 2! 2Fe(s) F2(g)  FeF3(s)

Pentane reacts with oxygen gas to produce carbon dioxide and water vapor. Chemical Equation C5H12 (g) + O2(g)  H2O(g) CO2(g) + Diatomic CnH(2n +2)

#8B 8 5 6 C5H12(g) + O2(g)  CO2(g) + H2O(g) C= H= O = 5 12 2 C= H =
pentane + oxygen  carbon dioxide + water 8 5 6 C5H12(g) + O2(g)  CO2(g) + H2O(g) C= H= O = 5 12 2 C= H = O = 5 2 11 1 2 3 5 12 16 5 12 16 C5H12(g) + 8O2(g)  5CO2(g) + 6H2O(g)

Solid magnesium hydroxide reacts with the gas ammonia to yield the gas magnesium nitride and liquid water Chemical Equation Memorize! Mg2+ OH- Mg2+ N3- Mg(OH)2 (s) + NH3(g)  Mg3N2(g) + H2O(l)

#9B Mg(OH)2(aq) + NH3(g)  Mg3N2(g) + H2O(l) 3 2 6 Mg= O = H = N = 3 6
magnesium hydroxide + ammonia  magnesium nitride + water Mg(OH)2(aq) NH3(g)  Mg3N2(g) + H2O(l) 3 2 6 Mg= O = H = N = 3 6 9 1 1 2 5 Mg = O = H = N = 3 6 12 2 3 1 2 3 6 12 2 3 Mg(OH)2(aq) NH3(g)  Mg3N2(g) H2O(l)

Oxygen gas plus the gas hydrogen chloride produces steam and chlorine gas Chemical Equation O2 (g) + HCl(g)  H2O(g) Cl2(g) + Diatomic! Diatomic!

#10B 2 O2(g) + HCl(g)  H2O(g) + Cl2(g) 2 4 2 2 Cl= H= O = 4 2 1 2 Cl=
oxygen + hydrogen chloride  water + chlorine 2 O2(g) + HCl(g)  H2O(g) + Cl2(g) 2 4 2 2 Cl= H= O = 4 2 1 2 Cl= H = O = 2 4 4 2 2 1 O2(g) + 4HCl(g)  2H2O(g) + 2Cl2(g)

Classifying Chemical Reactions
Section 3 Classifying Chemical Reactions

Study Model 4 complete Key Questions Set I #’s 1 - 6

Model 4 –I Class Check-In
Dissolved in water – aq Liquid – l Solid – s Gas - g 2. Sets C & D involve ions in solution 3. Sets A & B involve gases and/or solids 4. D Ionic compounds dissolved in water switch partners B One compound breaks into elements or smaller compounds. A Two or more elements or compounds combine to form one product. C Part of an ionic compound is removed and replaced by a new element.

Model 4 –I Class Check-In
5. Synthesis – Put several things together to make one. Decomposition – To break down or break something apart Displacement – the substitution of one item for another 6. Set A – Synthesis Reaction Set B – Decomposition Reaction Set C – Single Displacement Reaction Set D – Double Displacement

Study Model 4 complete Key Questions Set II #’s 7 - 12

Model 4 -II Class Check-In
7. Yes, two elements can be used as reactants for a synthesis reaction 4Fe(s) + 3O2(g)  2Fe2O3(s) 8. Yes, two compounds can be used as reactants for a synthesis reaction P2O5(g) + 3H2O(l)  2H3PO4(aq) 9. Elements and compounds can be seen in the products of decomposition reactions. MgCO3(s)  MgO(s) + CO2(g) 8Li2S(s)  16Li(s) + S8(s)

10. A single displacement reaction is the reaction between an element and a compound. A double displacement reaction is the reaction between two compounds dissolved in water; two aqueous solutions. 11.a. Write one synthesis reaction from Model 4 in reverse 2NH3(g)  N2(g) + 3H2(g) b. Label the reaction with one of the reaction types Decomposition

12. S DD D SD DD SD S D

Study Model 4 complete Key Questions Set III #’s 13 - 16

Model 4 -III Class Check-In
13. The correct formula for magnesium oxide is MgO b. The correct formula for elemental oxygen is O2 c. The first student has the incorrect formula for magnesium oxide, they forgot to cross charges and reduce, there should not be a subscript of a 2 after oxygen. d. The second student did not write the correct formula for oxygen, it is a diatomic molecule and should be written O2 e. The correctly balanced chemical equation is 2Mg(s) + O2(g)  2MgO(s)

14. Fluorine will displace bromide ion in lithium bromide; a halogen will be displaced by another halogen. The student formed FBr, he displaced lithium with fluorine, this compound is not an ionic compound and will not form in a single displacement reaction. c. The correctly balanced chemical equation is 2LiBr(aq) + F2(g)  Br2(g) + 2LiF(aq)

15. The student did not cross the charges to write the formulas, iron is a 3+ ion and would bind to three hydroxide ions. There should only be one bromide ion for every one sodium ion. b. FeBr3(aq) + 3NaOH(aq)  Fe(OH)3(s) +3NaBr(aq)

16. Diphosphorus pentoxide + water  phosphoric acid This is a synthesis reaction b. P2O5 + H2O  H3PO4 c. P2O5 + 3H2O  2H3PO4

Study Model 4 complete Key Questions Set IV # 17

Model 4 -IV Class Check-In
17. 2Al(s) + N2(g) 2AlN(s) Synthesis 2N2O(s)  2N2(g) + O2(g) Decomposition SrCl2(aq) + 2 AgNO3(aq)  2AgCl(s) + Sr(NO3)2(aq) Double Displacement d. 2Cr(NO3)3(aq) + 3ZnCl2(aq)  3Zn(NO3)2(aq) + 2CrCl3(s) Double Displacement

Model 4 -IV Class Check-In
17. e. 2Na(s) + Cl2(g) 2NaCl(s) Synthesis f. Zn(s) + 2HCl (aq)  H2(g) + ZnCl2(aq) Single Displacement

Study Model 5 complete Questions 18 - 23

Model 5Class Check-In 18. The products of all combustion reactions are gaseous carbon dioxide and water 19. The common reactant in all combustion reactions is elemental oxygen 20. A hydrocarbon is a compound that contains hydrogen and carbon 21. C5H12(g) + 8O2(g)  5CO2(g) + 6H2O(g) b. 2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(g)

Model 5Class Check-In 22. 2C3H7OH(g) + 9O2(g)  6CO2(g) + 8H2O(g)
C6H12O6(g) + 6O2(g)  6CO2(g) + 6H2O(g) 23. A combustion reaction can not be classified as: Synthesis – Combustion reactions have more than one product Decomposition – Combustion reactions have more than one reactant Single Displacement – The oxygen in a combustion reaction end up in both products, not just in one compound Double Displacement – Combustion reactions start with a compound and element, not two compounds.

Decomposition Reaction

Synthesis Reaction

Double Displacement

Single Displacement Reaction

Combustion

Problems C

Aqueous aluminum hydroxide yields aluminum oxide and water Chemical Equation Al(OH)3(aq)  Al2O3(s) + H2O (l) Balanced Chemical Equation Al2O3(s) + 3 H2O (l) 2 Al(OH)3(aq)  Type of Reaction Decomposition

Aqueous calcium fluoride reacts with aqueous hydrogen sulfate to give aqueous hydrogen fluoride and solid calcium sulfate. Chemical Equation CaF2(aq) HF(aq) + H2SO4(s)  + CaSO4(aq) Balanced Chemical Equation CaF2(aq) 2 H2SO4(s)  HF(aq) + + CaSO4(aq) Type of Reaction Double Displacement

Solid iron (III) oxide reacts with hydrogen gas to produce iron and water Chemical Equation Fe2O3(s) + H2O (l) H2(g)  Fe(s) + Balanced Chemical Equation Fe2O3(s) + 3H2O (l) 3H2(g)  2Fe(s) + Type of Reaction Single Displacement

Aluminum plus aqueous manganese (II) oxide yield manganese and aqueous aluminum oxide. Chemical Equation + MnO(aq) Mn(s) + Al2O3 (aq) Al(s)  Balanced Chemical Equation + 3 MnO(aq) 3 Mn(s) + Al2O3 (aq) 2 Al(s)  Type of Reaction Single Displacement

A solution of dihydrogen sulfide reacts with oxygen gas to produce water and aqueous sulfur dioxide. Chemical Equation H2S(aq) + SO2(aq) + O2(g)  H2O (l) Balanced Chemical Equation 2H2S(aq) + 2SO2(aq) + 3O2(g)  2H2O (l)

Solutions of beryllium iodide and tin(IV) nitrate yields aqueous beryllium nitrate and solid tin(IV) iodide Chemical Equation BeI2(aq) + Sn(NO3)4 (aq)  Be(NO3)2(aq) + SnI4(s) Balanced Chemical Equation 2 BeI2(aq) + Sn(NO3)4 (aq)  2 Be(NO3)2(aq) + SnI4(s) Type of Reaction Double Displacement

Solid iron reacts with fluorine gas to give solid iron (III) fluoride. Chemical Equation Fe(s) + F2(g) FeF3 (2)  Balanced Chemical Equation Fe(s) 2 + F2(g) FeF3 (2) 3  2 Type of Reaction Synthesis

Pentane reacts with oxygen gas to produce carbon dioxide and water vapor. Chemical Equation C5H12 (g) + O2(g)  H2O(g) CO2(g) + Balanced Chemical Equation C5H12 (g) + 8O2(g)  5CO2(g) 6H2O(g) + Type of Reaction Combustion

Solid magnesium hydroxide reacts with the gas ammonia to yield the gas magnesium nitride and liquid water Chemical Equation Mg(OH)2 (s) + NH3(g)  Mg3N2(g) H2O(l) + Balanced Chemical Equation 3 Mg(OH)2 (s) + 2 NH3(g)  Mg3N2(g) 6 H2O(l) +

Oxygen gas plus the gas hydrogen chloride produces steam and chlorine gas Chemical Equation O2 (g) + HCl(g)  H2O(g) Cl2(g) + Balanced Chemical Equation O2 (g) + 2 HCl(g)  H2O(g) Cl2(g) +

Complete Model 6 with your teacher

Model #6 Predicting Products
Synthesis A + X  AX metal oxide + Water  Metal hydroxide Calcium hydroxide  Ca2+ OH- + H2O(l)  Ca(OH)2(s) CaO(s) Solubility Rules – Oxides and hydroxides are not soluble

Synthesis A + X  AX CaO(s) + H2O(l)  Ca(OH)2(s) Ca= O= H = 1 2 Ca= H = O = 1 2 Balanced as written! CaO(s) + H2O(l)  Ca(OH)2(s)

Decomposition AX  A + X metal carbonate  Metal oxide + Carbon dioxide Potassium Carbonate + Carbon dioxide  Potassium oxide O2- K+ + CO2(g) K2CO3(s)  K2O(s) Solubility Rules

Decomposition AX  A + X K2CO3(s)  K2O(s) + CO2(g) K= C= O = 2 1 3 K= C = O = 2 1 3 Balanced as written! K2CO3(s)  K2O(s) + CO2(g)

Single Displacement A + BX  B + AX Check the Activity series Magnesium is more reactive than chromium Mg2+ Cl- Mg(s) + CrCl3(aq) Cr(s) + MgCl2(aq)  Solubility Rules

Single Displacement A + BX  B + AX Mg(s) CrCl3(aq)  Cr(s) MgCl2(aq) 3 2 2 2 3 1 ½ 1 2 Mg= Cr= Cl = 1 ½ 1 3 3 2 6 Mg= Cr= Cl = 1 3 3 2 6 2 6 Multiply all coefficients by 2 to get rid of the 1/2! 3Mg(s) + 2CrCl3(aq)  2Cr(s) MgCl2(aq)

Double Displacement AX + BY  BX + AY Ag+ Br- K+ NO3- KBr(aq) + AgNO3(aq) AgBr(s) + KNO3(aq)  Solubility Rules

Double Displacement AX + BY  BX + AY KBr(aq) + AgNO3(aq)  AgBr(s) + KNO3(aq) K= Br= Ag = NO3 = 1 1 K= Br = Ag = NO3 = Balanced as written! KBr(aq) + AgNO3(aq)  AgBr(s) + KNO3(aq)

Combustion Hydrocarbon + oxygen  carbon dioxide + water (Alkane)  Octane + oxygen Carbon dioxide + water CnH(2n+2) Diatomic CO2(g) + H2O(g) C8H18(g) +  O2(g)

Combustion Hydrocarbon + oxygen  carbon dioxide + water C8H18(g) O2(g)  CO2(g) H2O(g) 2 12 ½ 25 16 8 9 18 C= H= O = 8 18 25 1 2 3 8 2 17 C= H= O = 8 18 25 8 18 2 16 36 50 16 36 50 2C8H18(g) + 25 O2(g)  16O2(g) + 18 H2O(g) Multiply all coefficients by 2 to get rid of the 1/2!

Use Model 6 to complete Practice Problems on page 25

Model #6 Practice Problems
CoBr2(s)  Co(s) + Br2(g) Co= Br= 1 2 Co= Br= 1 2 Balanced as written! CoBr2(s)  Co(s) + Br2(g) Decompostion

2 K2SO4(aq) + Ba(NO3)2(aq)  BaSO4(s) + KNO3(aq) K= SO4= Ba = NO3 = 2 1 K= SO4 = Ba = NO3 = 2 1 1 K2SO4(aq) + Ba(NO3)2(aq)  BaSO4(s) + KNO3(aq) Double Displacement AX + BY  BX + AY

C5H12(g) O2(g)  CO2(g) H2O(g) 8 5 6 5 2 11 C= H= O = 5 12 16 1 2 3 C= H= O = 5 12 16 5 12 2 C5H12(g) O2(g)  CO2(g) H2O(g) Combustion Hydrocarbon + oxygen  carbon dioxide + water

K2O(s) + H2O(l)  KOH(s) 2 K= O= H = 2 K= H = O = 2 1 K2O(s) + H2O(l)  2KOH(s) Synthesis A + X  AX

Zn(s) HCl(aq)  H2(aq) ZnCl2(aq) 2 1 2 1 2 Zn= H= Cl = Zn= H= Cl = 1 Zn(s) HCl(aq)  H2(aq) ZnCl2(aq) Single Displacement A + BX  B + AX

Oh, That Curious Ira Remsen!

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