Answer
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:
$35.45g \times \frac{1L}{3.214g} = 11.03 L$
Chemistry (7th Edition)
by McMurry, John E.; Fay, Robert C.; Robinson, Jill Kirsten
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
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
