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: 3.94

Answer

Theoretical yield = 8.55 g Percent yield = 76.6%

Work Step by Step

1. Calculate or find the molar mass for $ S_8 $: $ S_8 $ : ( 32.07 $\times$ 8 )= 256.56 g/mol 2. Using the molar mass as a conversion factor, find the amount in moles: $$ 4.06 \space g \times \frac{1 \space mole}{ 256.56 \space g} = 0.01583 \space mole$$ 3. Calculate or find the molar mass for $ Cl_2 $: $ Cl_2 $ : ( 35.45 $\times$ 2 )= 70.90 g/mol 4. Using the molar mass as a conversion factor, find the amount in moles: $$ 6.24 \space g \times \frac{1 \space mole}{ 70.90 \space g} = 0.0880 \space mole$$ 5. Find the amount of product if each reactant is completely consumed. $$ 0.01583 \space mole \space S_8 \times \frac{ 4 \space moles \ S_2Cl_2 }{ 1 \space mole \space S_8 } = 0.0633 \space mole \space S_2Cl_2 $$ $$ 0.0880 \space mole \space Cl_2 \times \frac{ 4 \space moles \ S_2Cl_2 }{ 4 \space moles \space Cl_2 } = 0.0880 \space mole \space S_2Cl_2 $$ 6. Since the reaction of $ S_8 $ produces less $ S_2Cl_2 $ for these quantities, it is the limiting reactant. 7. Calculate or find the molar mass for $ S_2Cl_2 $: $ S_2Cl_2 $ : ( 35.45 $\times$ 2 )+ ( 32.07 $\times$ 2 )= 135.04 g/mol 8. Using the molar mass as a conversion factor, find the mass in g: $$ 0.0633 \space mole \times \frac{ 135.04 \space g}{1 \space mole} = 8.55 \space g$$ 9. Calculate the percent yield: $$percent \space yield = \frac{actual \space yield}{theoretical \space yield} \times 100\%$$ $$percent \space yield = \frac{6.55 \space g}{8.55 \space g} \times 100\% = 76.6\%$$
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