Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 3 - Chemical Reactions - Questions for Review and Thought - Topical Questions - Page 149e: 70a


$CO$ is the limiting reactant.

Work Step by Step

1. Calculate the number of moles of $H_2$: 1.01* 2 = 2.02g/mol $12g \times \frac{1 mol}{ 2.02g} = 5.94mol (H_2)$ - The balanced reaction is: $CO + 2H_2 -- \gt CH_3OH$ The ratio of $H_2$ to $CO$ is 2 to 1: $5.94 mol (H_2) \times \frac{ 1 mol(CO)}{ 2 mol (H_2)} = 2.97mol (CO)$ 2. Calculate the mass of $CO$: 12.01* 1 + 16* 1 = 28.01g/mol $2.97 mol \times \frac{ 28.01 g}{ 1 mol} = 83.2g (CO)$ The necessary mass of $CO$ is higher than 74.5g Therefore, $CO$ is the limiting reactant.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After 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.