Chem 1211 1st Edition Lecture 11 Outline of Last Lecture I Oxidation Reduction Reactions II Stoichiometry Outline of Current Lecture I Stoichiometry Continued II Limiting Reagents III Info for Stoichiometry Current Lecture I Stoichiometry Continued a ConcepTest For the reaction Cl2 g H2 g 2 HCl g how many grams of HCl are produced from 7 09 grams of chlorine gas i ANSWER 7 29 grams ii Reasoning convert 7 09 grams Cl2 to grams HCl by multiplying 7 09g Cl2 x 1 mol Cl2 70 9 g Cl2 x 2 mol HCl 1 mol Cl2 x 36 458 g HCl 1 mol HCl b ConcepTest For the reaction 4 NH3 5 O2 4 NO 6 H2O if 17 g of ammonia reacts how many moles of hydrogen atoms are contained in the products i ANSWER 3 moles of H ii 17g NH3x 1 mol 17g NH3 x 6 mol H2O 4 mol NH3 x 2 mol H 1 mole H2 c Calculation Strategy i Mass of moles use formula weight ii Ration of moles use stoichiometric coefficients iii Moles to mass use formula weight d ConcepTest Consider the reaction Zn 2 HCl ZnCl2 H2 If 65 4 g of Zn reacts with 36 5 g of HCl what is the total mass of the ZnCl2 and H2 i ANSWER 692 g ii For every one Zinc you need two HCl Since we only have one mole of HCl only half of the zinc will react This is an example of a limiting reagent These notes represent a detailed interpretation of the professor s lecture GradeBuddy is best used as a supplement to your own notes not as a substitute II III Limiting Reagents a ConcepTest Consider the reaction Zn 2 HCl ZnCl2 H2 Which case has zinc as the limiting reaction i 1 mol Zn 1 mol HCl ii 1 5 mol Zn 2 mol HCl iii 5 mol Zn 2 mol HCl iv 2 mol Zn 1 mol HCl v 1 mol Zn 2 mol HCl 1 ANSWER iii 5 mol Zn 2 mol HCl 2 Reasoning You need a 1 2 ratio but iii has a 1 4 ratio Info for Stoichiometry a Some information comes from experiments i Formula ii Ratios of reagents b Quantitative analysis chemical analysis use known reaction known amounts of one substance to determine another c Combustion analysis burn known amount and analyze carbon hydrogen and other atoms d ConcepTest Which empirical formula corresponds to a molecule that has the formula weight of 88 i CH ii CH2 iii CH2O iv C2H4O v None of these 1 ANSWER C2H4O
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