Chy102-Tutorial 04

Chy102-Tutorial 04 - 1 General Chemistry Chem. 102 for...

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Unformatted text preview: 1 General Chemistry Chem. 102 for Engineering General Chemistry General Chemistry Chem. 102 for Engineering Chem. 102 for Engineering Tutorial 4 2 Objectives 7 Types of chemical reactions occurring in aqueous solutions ¾ Precipitation reactions ¾ Acid-base reactions ¾ Reduction-oxidation reactions 3 A student reacted 11.3 g ethanol (46.05) and 3.48 g PCl 3 (137.32). The actual yield of C 2 H 5 Cl was 1.24 g. Calculate the percent yield of C 2 H 5 Cl (64.51). 3C 2 H 5 OH + PCl 3 3C 2 H 5 Cl + H 3 PO 3 a) 35.6% C 2 H 5 Cl b) 8.39% C 2 H 5 Cl c) 5.99% C 2 H 5 Cl d) 25.3% C 2 H 5 Cl e) 7.85% C 2 H 5 Cl Solution 3 3 5 2 3 5 2 PO H + Cl H C 3 PCl + OH H C 3 → 11.3 g + 3.48 g ? g 0.245 moles 0.025 moles moles 05 . 46 3 . 11 moles 32 . 137 48 . 3 67 . 9 = 025 . 245 . = PCl mol OH H C mol = present actually ratio Mole 3 5 2 3 = 1 3 = PCl mol OH H C mol = required ratio Mole 3 5 2 Molar mass of C 2 H 5 OH = 46.05 Molar mass of PCl 3 = 137.32 Since the actual ratio is more than 3, PCl 3 is the limiting reactant 3 5 2 3 5 2 mol 075 . = PCl mol 1 Cl H C mol 3 × PCl mol 025 . = formed Cl H C of moles of . No Cl H C g 84 . 4 = Cl H C mol 1 Cl H C g 51 . 64 × Cl H C mol 075 . = formed Cl H C of mass 5 2 5 2 5 2 5 2 5 2 % 63 . 25 = g 84 . 4 g 24 . 1 = ) grams ( yield l theoretical ) grams ( yield actual = Cl H C of yield percent 5 2 3 3 5 2 3 5 2 PO H + Cl H C 3 PCl + OH H C 3 → 11.3 g + 3.48 g ? g 0.245 moles 0.025 moles 67 . 9 = 025 . 245 . = PCl mol OH H C mol = present actually ratio Mole 3 5 2 3 = 1 3 = PCl mol OH H C mol = required ratio Mole 3 5 2 Molar mass of C 2 H 5 OH = 46.05 Molar mass of PCl 3 = 137.32 Since the actual ratio is more than 3, PCl 3 is the limiting reactant Cl H C g 84 . 4 = Cl H C mol 1 Cl H C g 51 . 64 × Cl H C mol 075 . = formed Cl H C of mass 5 2 5 2 5 2 5 2 5 2 Cl H C g 84 . 4 = Cl H C mol 1 Cl H C g 51 . 64 × Cl H C mol 075 . = formed Cl H C of mass 5 2 5 2 5 2 5 2 5 2 4 Precipitation Reactions 9 If two soluble substances (call them AX and BZ) are combined, you can assume that the products will be AZ and BX: AX( aq ) + BZ( aq ) AZ + BX Your goal is to determine, based on your knowledge of solubility rules, whether AZ or BX will form a solid (precipitate)....
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This note was uploaded on 12/13/2010 for the course CHEM 102 taught by Professor Raafat during the Spring '10 term at Ain Shams University.

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Chy102-Tutorial 04 - 1 General Chemistry Chem. 102 for...

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