Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises - Page 117: 3.69a

Answer

About 0.0055 moles of aluminum.

Work Step by Step

1. Find the volume of aluminum: $V = 1.00cm^2 * 0.0550cm$ $V = 0.055cm^3$ 2. Find the mass of aluminum, using the density: $d = m\div v$ $2.699g/cm^3 = m \div 0.055cm^3$ $2.699 \times 0.055 = m$ $m \approx 0.1484g $ 3. Find the number of moles of aluminum: $mm = m \div nºmoles$ $26.98 = 0.1484 \div nºmoles$ $nºmoles = 0.1484 \div26.98$ $nºmoles \approx 0.0055 $
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.