PracticeSheet13 key

# PracticeSheet13 key - Practice Sheets 13 BLC & DLC...

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Unformatted text preview: Practice Sheets 13 BLC & DLC Answers 13B.1 First calculate the theoretical yield, and then do the simple division. (20.0 g Al) 3 3 3 AlBr mol 1 AlBr g 266.69 Al mol 1 AlBr mol 1 Al g 26.98 Al mol 1 = 198 g Al Percent yield = Al g 198 Al g 87.4 × 100 = 44.2 % yield 13B.2 a) (1 mol Fe) Fe mol 4 O mol 3 2 = 0.75 mol O 2 . We have excess O 2 , so Fe is limiting. Alternatively, you could start with one mole of O 2 and determine that you don’t have enough iron to react with all of the oxygen. b) (7.5 mol Fe) Fe mol 4 O mol 3 2 = 5.6 mol O 2 . We have excess O 2 , so Fe is limiting. c) (160.0 g Fe) Fe mol 4 O mol 3 Fe g 55.85 Fe mol 1 2 = 2.149 mol O 2 . We have only 2 mol O 2 , so O 2 is limiting ....
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## This note was uploaded on 11/04/2011 for the course CHEM 105 taught by Professor Macedone during the Fall '07 term at BYU.

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PracticeSheet13 key - Practice Sheets 13 BLC & DLC...

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