# Chapter 3 - textbook answers - (the oxygen from the H 2 O =...

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Chapter 3 4. 0.115 C 7 H 16 + X O 2 Y CO + Z CO 2 + 0.92 H 2 O 11.5 g / 100 g/mol = 0.115 mol 8 x 0.115 = 0.92 1.3 – leftover O 2 = X 7 x 0.115 = 0.805 = Y + Z 1.05 mol = leftover O 2 + Y + Z 1.05 – 0.805 = 0.245 = leftover O 2 1.3 – 0.245 = 1.055 O 2 used For all the oxygens: 2 x 1.055 = 2.11 = Y + 2 Z + 0.92

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Unformatted text preview: (the oxygen from the H 2 O = 0.92) Using 0.805 = Y + Z (from earlier) 2.11 – 0.805 = Z + 0.92 so Z = 0.385 Solving for X and Y offers the rest of the answers a) 0.42 mol b) 0.385 mol c) 0.245 mol + --------------1.05 mol total...
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## This note was uploaded on 02/21/2011 for the course CHE 144 taught by Professor Kerber during the Fall '10 term at SUNY Stony Brook.

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Chapter 3 - textbook answers - (the oxygen from the H 2 O =...

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