1) C + H 2 → C 3 H 8 2) C 6 H 12 + O 2 → H 2 O + CO 2 3) NaI + Pb(SO 4 ) 2 → PbI 4 + Na 2 SO 4 4) HgI 2 + O 2 → HgO + I 2 5)List the 7 diatomic molecules.

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1) C + H 2 → C 3 H 8 2) C 6 H 12 + O 2 → H 2 O + CO 2 3) NaI + Pb(SO 4 ) 2 → PbI 4 + Na 2 SO 4 4) HgI 2 + O 2 → HgO + I 2 5)List the 7 diatomic molecules. Warm-up

Predict Products and balance: – MgO → – 2MgO → 2Mg + O 2 – CaCO 3 → – CaCO 3 → CaO + CO 2 – Na + O 2 → – 4Na + O 2 → 2Na 2 O

Combustion and Single Replacement Predicting Products, Day Four Whitaker 30 October 2014

Combustion Products will always be CO 2 and H 2 O Look for a carbon containing compound and O 2 as a reactants. The hardest work you have to do is balance the equation correctly.

Practice CH 4 + O 2  – CH 4 + 2O 2  CO 2 + 2H 2 O Cs + F 2  – 2Cs + F 2  2CsF Ba + Cl 2  – Ba + Cl 2  BaCl 2 C 3 H 8 + O 2  – C 3 H 8 + 5O 2  3CO 2 + 4H 2 O

Single Replacement What is a single replacement reaction? One “kicks” out the less reactive element. – Ex. 2Na (s) +2HCl (aq)  2NaCl (aq) + H 2(g)

Reactivity Elements range in reactivity – Some are very stable as elements These are less reactive Example: Gold (Au) – Some are very reactive These are mostly found in compounds (not by themselves) Example: Sodium (Na)

Activity Series The most active elements are at the top of the series An elemental metal can replace any ion that is BELOW it on the series If an element is more reactive, it will replace the less reactive ion in a compound.

Activity Series How to use the activity series. 1.Is your solo element a metal or a halogen? Which will it replace in the compound? 2.Find the Solo element in the correct activity series and compare it to what it wants to replace 3.If the solo element is above then the reaction will take place. If not, write “no reaction”

Al (s) + ZnCl 2(aq)  To replace the Zinc, Aluminum must be higher on the series Cu (s) + NaCl (aq)  Can copper replace sodium in the compound? Zn (s) + AlCl 3(aq) NO REACTION

Practice! Cr (s) + Pb(NO 3 ) 2(aq)  Pt (s) + CaCl 2(aq)  Ca (s) + FeO (aq)  Cr(NO 3 ) 2(aq) + Pb (s) NO REACTION CaO (aq) + Fe (s)

Warm-up Predict the products and balance! Ca + FeO  C 2 H 6 + O 2  Cs + F 2 

Warm-up Predict the products and balance! F 2 + AgCl  CH 2 + O 2  NaOH  Zn + O 2 

Double Replacement Unit 6, Day 5 Whitaker 3 November 2014

Remember Double Replacement Anions switch places and are each bonded to a different cation AB + CD  AD + CB The cation always has to be put before the anion.

Solubility rules All double replacement reactions will produce a precipitate If no precipitate is formed, then “no reaction” occurs – A precipitate is a solid that’s produced during a chemical reaction in a solution

Solubility If something is soluble, then it can be dissolved by water and is aqueous If something is insoluble, then it cannot be dissolved in water and is solid One product must be solid or no reaction occurs

Solubility Rules: Soluble Soluble – All Nitrates, Acetates, Ammoniums, and Group 1 salts. – All Chlorides, Bromides, and Iodides, except Silver, Lead, and Mercury (I) – All Fluorides except Group 2, Lead (II), and Iron (III) – All Sulfates except Calcium, Strontium, Barium, Mercury, Lead(II), and Silver

Solubility: Insoluble Insoluble – All Carbonates and Phosphates except Group 1 and Ammonium – All Hydroxides except Group 1, Strontium, Barium, and Ammonium – All Sulfides except Group 1, Group 2, and Ammonium – All Oxides except Group 1

Steps 1.First break the reactants into their ions (find the charges!). 2.Next, swap and drop partners for both (use reference table!) 3.Check solubility rules to see if a solid (precipitate) has formed. 4.Write complete balanced equation with states of matter.

Example Sodium Hydroxide + Copper (II) Sulfate

Predict the products and determine if a precipitate forms. Sodium phosphate + Nickel (II) chloride – NaCl and Ni 3 (PO 4 ) 2. Lead (II) Nitrate + Potassium Iodide – PbI 2 and KNO 3 Sodium Hydroxide + Potassium Chloride – NaCl and KOH Sodium phosphate + Lead (IV) nitrate – Pb 3 (PO 4 ) 4 and NaNO 3

Ba(NO 3 ) 2 (aq) + K 2 SO 4 (aq) → AgNO 3 (aq) + NaBr (aq) →

FeCl 3 (aq) + KOH (aq) → Pb(NO 3 ) 2 (aq) + K 2 SO 4 (aq) → Cu(NO 3 ) 2 (aq) + NaOH →

Writing the Net Ionic equation You almost know how to do this! 1.Start with the completely balanced equation. 2.Look at the solid product and make it the product of your Net Ionic equation. 3.For the reactants, put the ions that lead to the product

Example Na 2 SO 4 + CaCl 2  2NaCl + CaSO 4 – SO 4 2- (aq) + Ca 2+ (aq)  CaSO 4 (s) 3NaOH + FeCl 3  3NaCl + Fe(OH) 3 – 3OH - (aq) + Fe 3+ (aq)  Fe(OH) 3 (s)