Chemistry (7th Edition)

Published by Pearson
ISBN 10: 0321943171
ISBN 13: 978-0-32194-317-0

Chapter 1 - Chemical Tools: Experimentation and Measurement - Section Problems - Page 31: 84


I would need 11.2 L of gaseous hydrogen and 11.03 L of gaseous chlorine.

Work Step by Step

We use unit conversions to find: Hydrogen: $1.0078g \times \frac{1L}{0.0899g} = 11.2 L$ Chlorine: $35.45g \times \frac{1L}{3.214g} = 11.03 L$
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