Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 3 - Mass Relationships in Chemical Reactions - Questions & Problems: 3.65

Answer

3.6 mol $CO_{2}$

Work Step by Step

$2CO(g)+0_{2}(g)$ -> $2CO_{2}(g)$ mol of $CO_{2}$= $\frac{(3.6 mol CO)\times(2 mol CO_{2})}{(2 mol CO)}$ = 3.6 mol $CO_{2}$
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