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 - Page 111: 3.91

Answer

The percent yield of this reaction is 87.2%

Work Step by Step

1,. Determine the molar masses: $ FeTiO_3 $ : ( 47.87 $\times$ 1 )+ ( 55.85 $\times$ 1 )+ ( 16.00 $\times$ 3 )= 151.72 g/mol $$ \frac{1 \space mole \space FeTiO_3 }{ 151.72 \space g \space FeTiO_3 } \space and \space \frac{ 151.72 \space g \space FeTiO_3 }{1 \space mole \space FeTiO_3 }$$ $ TiO_2 $ : ( 47.87 $\times$ 1 )+ ( 16.00 $\times$ 2 )= 79.87 g/mol $$ \frac{1 \space mole \space TiO_2 }{ 79.87 \space g \space TiO_2 } \space and \space \frac{ 79.87 \space g \space TiO_2 }{1 \space mole \space TiO_2 }$$ 2. Find the theoretical amount of $TiO_2$: $$ 8.00 \times 10^{3} \space g \space FeTiO_3 \times \frac{1 \space mole \space FeTiO_3 }{ 151.72 \space g \space FeTiO_3 } \times \frac{ 1 \space mole \space TiO_2 }{ 1 \space mole \space FeTiO_3 } \times \frac{ 79.87 \space g \space TiO_2 }{1 \space mole \space TiO_2 } = 4.21 \times 10^{3} \space g \space TiO_2 $$ 3. Calculate percent yield: $$\frac{ 3.67 \times 10^3 \space g \space TiO_2 }{ 4.21 \times 10^{3} \space g \space TiO_2 } \times 100\% = 87.2 \%$$
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