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+ Synthesis/Combination Decomposition Single-Replacement  Activity Series Double-Replacement  Solubility Rules Combustion O2O2 ++ hydrocarbon CO 2 H2OH2OO2O2.

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Presentation on theme: "+ Synthesis/Combination Decomposition Single-Replacement  Activity Series Double-Replacement  Solubility Rules Combustion O2O2 ++ hydrocarbon CO 2 H2OH2OO2O2."— Presentation transcript:

1 + Synthesis/Combination Decomposition Single-Replacement  Activity Series Double-Replacement  Solubility Rules Combustion O2O2 ++ hydrocarbon CO 2 H2OH2OO2O2

2 Metals Lithium Potassium Calcium Sodium Magnesium Aluminum Zinc Chromium Iron Nickel Tin Lead Hydrogen* Copper Mercury Silver Platinum Gold Halogens Fluorine Chlorine Bromine Iodine Decreasing Activity

3

4 3.28 a) What is the mass of 2.50 x mol of MgCl 2 ? FW(MgCl 2 ) = Mg + 2 Cl = g/mol + 2( g/mol) FW(MgCl 2 ) = g/mol Mass = (2.50 x mol)( g/mol) = 2.38 g b) How many moles of NH 4 Cl are there in 75.6 g of this substance? FW(NH 4 Cl) = g/mol Moles = mass/FW = 76.5 g/( g/mol) = 1.43 mol FW = Formula Weightgfm = gram formula massMW = Molar Weight These are all equivalent terms for the molar mass of a substance:

5 c) How many molecules are there in mol HCHO 2 ? #m’cules = (6.022 x )( mol) = 4.65 x m’cules d) How many nitrate ions, NO 3 , are there in 4.88 x 10  3 mol Al(NO 3 ) 3 ? #f.u. = (6.022 x )(4.88 x 10  3 mol) = x f.u. #ions = (2.939 x f.u.)(3 NO 3  /f.u.) = 8.82 x ions Remember that ionic compounds don’t exist as discrete molecules, but are instead an extended lattice of ions. Hence, they are described in terms of the number of formula units (f.u.’s) that are present. In one f.u. of aluminum nitrate, there are three nitrate ions.

6 Perchlorate Perbromate Periodate ClO 4  BrO 4  IO 4  NitrateNO 3  Chlorate Bromate Iodate ClO 3  BrO 3  IO 3  Sulfate Carbonate SO 4 2  CO 3 2  NitriteNO 2  Chlorite Bromite Iodite ClO 2  BrO 2  IO 2  SulfiteSO 3 2  Hypochlorite Hypobromite Hypoiodite ClO  BrO  IO  Phosphate Arsenate PO 4 3- AsO 4 3- Polyatomic ions that you need to know are: The ‘ides’ Peroxide O 2 2  Hydroxide OH  Cyanide CN  Acetate C 2 H 3 O 2  OR CH 3 COO  The halogenates Permanganate MnO 4  Dichromate Cr 2 O 7 2  Chromate CrO 4 2  Form colored solutions The only cation Ammonium NH 4 +

7 BrO 4 , +7 SO 4 2 , +6 BrO 3 , +5 BrO 2 , +3 BrO , +1 SO 3 2 , +4 S 2 O 3 2 , +6 ClO 4 , +7 ClO 3 , +5 ClO 2 , +3 ClO , +1 IO 4 , +7 IO 3 , +5 IO 2 , +3 IO , +1 NO 3 , +5 NO 2 , +3 PO 4 3 , +5 PO 3 3 , +3 NH 4 +, -3 CrO 4 2 , +6 Cr 2 O 7 2 , +6 CO 3 2 , +4 C2H3O2C2H3O2

8 Combustion Analysis 3.48 (a) The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. Combustion of 2.78 mg of ethyl butyrate produces 6.32 mg of CO 2 and 2.58 mg of H 2 O. What is the empirical formula of the compound (element order is C H O)? C x H y O z + O 2 CO 2 + H 2 O 2.78 mg6.32 mg2.58 mg Step 1: Determine the number of moles of C and H atoms FW(CO 2 ) = mg/mmolFW(H 2 O) = mg/mmol mmoles C = (6.32 mg CO 2 )/( mg/mmol) = mmol C mmoles H 2 O = (2.58 mg H 2 O)/( mg/mmol) = mmol mmoles H = ( mmoles H 2 O)(2 H atoms/H 2 O) = mmol H

9 Step 2: Convert the mmoles of C and H to mg of the two elements so can determine how much oxygen was in the compound. mg C = ( mmol)( mg/mmol) = mg C mg H = ( mmol)( mg/mmol) = mg H mg O = 2.78 mg – ( mg mg) = mg O mmol O = (0.766 mg)/( mg/mmol) = mmol Step 3: Divide through by the smallest molar amount to determine the empirical formula. C: mmol/ mmol = 3 H: mmol/ mmol = 6 O: mmol/ mmol = 1 The empirical formula is C 3 H 6 O

10 Stoichiometry 3.56 The fermentation of glucose (C 6 H 12 O 6 ) produces ethyl alcohol (C 2 H 5 OH) and CO 2. (a) How many moles of CO 2 are produced when mol of C 6 H 12 O 6 reacts in this fashion? C 6 H 12 O 6 (aq)  2 C 2 H 5 OH (aq) + 2 CO 2 (g) mol #mol x2x mol (b) How many grams of C 6 H 12 O 6 are needed to form 7.50 g of C 2 H 5 OH? mol C 2 H 5 OH = (7.50 g)/( g/mol) = mol FW(C 2 H 5 OH) = g/molFW(C 6 H 12 O 6 ) = g/mol mol mol Mass glucose = ( mol)( g/mol) = 14.7 g

11 LR: 3.70 Aluminum hydroxide reacts with sulfuric acid as shown below. (b) How many moles of Al 2 (SO 4 ) 3 form when mol Al(OH) 3 and mol H 2 SO 4 are allowed to react ? Limiting Reactants/Theoretical Yield 2 Al(OH) 3 (s) + 3 H 2 SO 4 (aq)  Al 2 (SO 4 ) 3 (aq) + 6 H 2 O (l) mol mol 2x3xx mol mol The H 2 SO 4 is the limiting reactant because it limits how much aluminum sulfate will be formed. The theoretical (calculated) yield for this reaction is mol of aluminum sulfate and 1.10 mol of water. What is the percent yield for this reaction if only mol of water are collected? %yield = (0.750 mol)/(1.10 mol)*100 = 68.2% 6x


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