Chemistry: Principles and Practice (3rd Edition)

Published by Cengage Learning
ISBN 10: 0534420125
ISBN 13: 978-0-53442-012-3

Chapter 3 - Equations, the Mole, and Chemical Formulas - Questions and Exercises - Exercises - Page 132: 3.47

Answer

$Zn (s) + HCl (l) → ZnCl_{2} (s) + H_{2} (g)$ Oxidation numbers in reactants: Zn = 0, In HCl, H = +1, Cl = -1 Oxidation numbers products: ZnCl2: Zn = +2, Cl = -1, H = 0 Zn is oxidized and hydrogen is reduced

Work Step by Step

$Zn (s) + HCl (l) → ZnCl_{2} (s) + H_{2} (g)$ Oxidation numbers in reactants: Zn = 0, In HCl, H = +1, Cl = -1 Oxidation numbers products: ZnCl2: Zn = +2, Cl = -1, H = 0 Zn is oxidized and hydrogen is reduced
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