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: 3.20c

Answer

$2C_3H_7OH (l) + 9O_2 (g) \rightarrow 6CO_2 (g) + 8H_2O (l)$

Work Step by Step

In every combustion reaction, $CO_2$ and water products. The balanced equation is provided in the answer, but an effective method to balanced chemical equations would be to count the number of molecules for each element on both sides of the reaction. Only change the coefficient and not the subscript. We start with the given unbalanced form: $C_3H_7OH (l) + O_2 (g) \rightarrow CO_2 (g) + H_2O (l)$ Left: C:3, H:8,O:3 Right: C:1,H2,O:3 $C_3H_7OH (l) + O_2 (g) \rightarrow 3CO_2 (g) + H_2O (l)$ Left: C:3, H:8,O:3 Right: C:3,H:2,O:7 $C_3H_7OH (l) + O_2 (g) \rightarrow 3CO_2 (g) + 4H_2O (l)$ Left: C:3, H:8,O:3 Right: C:3,H:8,O:7 $C_3H_7OH (l) + 2O_2 (g) \rightarrow 3CO_2 (g) + 4H_2O (l)$ Left: C:3, H:8,O:5 Right: C:3,H:8,O:7 $2C_3H_7OH (l) + 2O_2 (g) \rightarrow 3CO_2 (g) + 4H_2O (l)$ Left: C:6, H:16,O:6 Right: C:3,H:8,O:6 $2C_3H_7OH (l) + 2O_2 (g) \rightarrow 6CO_2 (g) + 4H_2O (l)$ Left: C:6, H:16,O:6 Right: C:6,H:8,O:13 $2C_3H_7OH (l) + 2O_2 (g) \rightarrow 6CO_2 (g) + 8H_2O (l)$ Left: C:6, H:16,O:6 Right: C:6,H:16,O:20 $2C_3H_7OH (l) + 9O_2 (g) \rightarrow 6CO_2 (g) + 8H_2O (l)$ Left: C:6, H:16,O:20 Right: C:6,H:16,O:20
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.