Chemistry: The Central Science (13th Edition)

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

Chapter 5 - Thermochemistry - Exercises - Page 208: 5.80a


$1CH_3OH(l) + \frac{3}{2}O_2(g) --> 1CO_2(g) + 2H_2O(g)$

Work Step by Step

1. Write the unbalanced equation: $CH_3OH(l) + O_2(g) --> CO_2(g) + H_2O(g)$ 2. Balance the number of Carbon moles: $1CH_3OH(l) + O_2(g) --> 1CO_2(g) + H_2O(g)$ 3. Balance the number of Hydrogen moles: $1CH_3OH(l) + O_2(g) --> 1CO_2(g) + 2H_2O(g)$ 4. Balance the number of Oxygen moles: $1CH_3OH(l) + \frac{3}{2}O_2(g) --> 1CO_2(g) + 2H_2O(g)$ *Since we got 4 Oxygen moles in the right side, and we already have 1 in $1CH_3OH(l)$, we can put a $(\frac{3}{2})$ next to the $O_2$ and the equation will be totally balanced.
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