Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 10 - Gases - Exercises - Page 436: 10.57a



Work Step by Step

The formula for glucose is: C$_6$H$_12$O$_6$ + 6O$_2$ --> 6CO$_2$ + 6H$_2$O We are given 24.5g of glucose at T = 37 Celsius = 310 Kelvin P = 0.970 atm And we want to find the volume of CO$_2$ produced when 24.5 of glucose is consumed. We know from the formula that for 1 mole of glucose consumed, 6 moles of CO$_2$ are produced so the ratio is; 1 mole glucose : 6 mole CO$_2$ 24.5 g glucose x (1 mol glucose/180g glucose) = 0.136 mole 1 mole glucose : 6 mole CO$_2$ = 0.136 mole glucose : x mole CO$_2$ By using simple algebra, we get x = 0.816 mole CO$_2$ produced Now to find the volume of this amount of CO$_2$ using the combined gas law, V = nRT/P V = (0.816 mol x 0.08201 x 310)/(0.970) = 21.4L
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