Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 3 - Mass Relationships in Chemical Reactions - Questions & Problems - Page 110: 3.72


$201.9g$ ethanol volume = $256L$

Work Step by Step

$C_{6}H_{12}O_{6}$ -> $2C_{2}H_{5}OH+2CO_{2}$ Mass of ethanol = $\frac{(500.4gGlucose)\times(1molGlucose)\times(2molEthanol)\times(46.08gEthanol)}{(180.18gGlucose)\times(1molEthanol)\times(1molEthanol)}$ $\approx$ 201.9436436 =201.9 g volume = $\frac{201.9g}{0.789g/L}$ $=256L$
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