General, Organic, and Biological Chemistry: Structures of Life (5th Edition)

Published by Pearson
ISBN 10: 0321967461
ISBN 13: 978-0-32196-746-6

Chapter 7 - Chemical Reactions and Quantities - 7.8 Limiting Reactants and Percent Yield - Questions and Problems - Page 271: 7.83

Answer

This reaction can produce 70.9 g of $Al_2O_3$.

Work Step by Step

$ Al $ : 26.98 g/mol $$ \frac{1 \space mole \space Al }{ 26.98 \space g \space Al } \space and \space \frac{ 26.98 \space g \space Al }{1 \space mole \space Al }$$ $ Al_2O_3 $ : ( 26.98 $\times$ 2 )+ ( 16.00 $\times$ 3 )= 101.96 g/mol $$ \frac{1 \space mole \space Al_2O_3 }{ 101.96 \space g \space Al_2O_3 } \space and \space \frac{ 101.96 \space g \space Al_2O_3 }{1 \space mole \space Al_2O_3 }$$ $$ 50.0 \space g \space Al \times \frac{1 \space mole \space Al }{ 26.98 \space g \space Al } \times \frac{ 2 \space moles \space Al_2O_3 }{ 4 \space moles \space Al } \times \frac{ 101.96 \space g \space Al_2O_3 }{1 \space mole \space Al_2O_3 } = 94.5 \space g \space Al_2O_3 $$ $$actual \space yield =\frac{ percent \space yield \times theoretical \space yield }{100\%}$$$$actual \space yield =\frac{( 75.0 \%)\times ( 94.5 \space g \space Al_2O_3 )}{100\%} = 70.9 \space g \space Al_2O_3 $$
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.