Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 3 - Chemical Reactions - Questions for Review and Thought - Topical Questions - Page 148d: 44c


- This is a gas formation reaction. Products: $2KClO_4$ $CO_2$ and $H_2O$ Completed equation: $K_2CO_3(aq) + 2HClO_4(aq) \lt -- \gt 2KClO_4(aq) + CO_2(g) + H_2O(l)$

Work Step by Step

1. Write and analyze the products of the reaction: - Since we have 2 ionic compounds, we will have a double displacement reaction. So, the products are: $KClO_4$: Which is soluble in water* $H_2CO_3$: Which is soluble in water* * According to table 3.1 (Page 101). Since we have an insoluble product, there will be precipitation. - But, the carbonic acid $(H_2CO_3)$, turns to $CO_2(g)$ and $H_2O(l)$. - Therefore, this is a gas formation reaction. Since we already have the reactants and the products, the chemical reaction is: $K_2CO_3(aq) + HClO_4(aq) \lt -- \gt KClO_4(aq) + CO_2(g) + H_2O(l)$ - Balancing the reaction: $K_2CO_3(aq) + 2HClO_4(aq) \lt -- \gt 2KClO_4(aq) + CO_2(g) + H_2O(l)$
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