Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 5 - Gases - Exercises - Page 237: 65

Answer

$3.21\text{ grams Al}$

Work Step by Step

$2.00 \text{ text L $O_2$}*\frac{1\text{ mol}}{22.42\text{ L}}=0.0892\text{ mol $O_2$}$ $0.0892\text{ mol $O_2$}*\frac{4\text{ mol Al}}{3\text{ mol $O_2$}}*\frac{26.981538\text{ grams Al}}{1\text{ mol Al}}=3.21\text{ grams Al}$
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