Chemistry: The Central Science (13th Edition)

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

Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises - Page 113: 3.14d

Answer

$4C_2H_5NH_2(g)+15O_2(g)\rightarrow 8CO_2(g)+14H_2O(g)+2N_2(g)$

Work Step by Step

$C_2H_5NH_2(g)+O_2(g)\rightarrow CO_2(g)+H_2O(g)+N_2(g)$ Add a coefficient of 2 to $C_2H_5NH_2$ to create an even number of $H$ atoms on the left. Since $H$ only appears in pairs on the right, the number we are balancing to must be even. $2C_2H_5NH_2(g)+O_2(g)\rightarrow CO_2(g)+H_2O(g)+N_2(g)$ Add a coefficient of 4 to $CO_2$ to balance $C$. $2C_2H_5NH_2(g)+O_2(g)\rightarrow 4CO_2(g)+H_2O(g)+N_2(g)$ There are now 14 $H$ on the left and only 2 on the right. Add a coefficient of 7 to $H_2O$ to balance $H$. $2C_2H_5NH_2(g)+O_2(g)\rightarrow 4CO_2(g)+7H_2O(g)+N_2(g)$ There are now an odd number of $O$ atoms on the right. $O$ only appears in pairs on the left side. Double all coefficients to elimnate the negative. $4C_2H_5NH_2(g)+2O_2(g)\rightarrow 8CO_2(g)+14H_2O(g)+2N_2(g)$ There are now 30 $O$ atoms on the left and only 4 on the left. Increase the coefficient of $O_2$ from 2 to 15 to provide the 16 additional $O$ atoms. $4C_2H_5NH_2(g)+15O_2(g)\rightarrow 8CO_2(g)+14H_2O(g)+2N_2(g)$
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