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Ch.3 - Chemical Reactions and Reaction Stoichiometry
Chapter 3, Problem 83b

When benzene 1C6H62 reacts with bromine 1Br22, bromobenzene 1C6H5Br2 is obtained: C6H6 + Br2¡C6H5Br + HBr (b) If the actual yield of bromobenzene is 42.3 g, what is the percentage yield?

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insert step 1: Begin by writing the balanced chemical equation for the reaction: C_6H_6 + Br_2 \rightarrow C_6H_5Br + HBr.
insert step 2: Calculate the molar mass of C_6H_6 (benzene) and C_6H_5Br (bromobenzene) using the periodic table.
insert step 3: Determine the theoretical yield of C_6H_5Br by using stoichiometry. Start with the amount of C_6H_6 you have, convert it to moles, and use the balanced equation to find the moles of C_6H_5Br that should be produced.
insert step 4: Convert the moles of C_6H_5Br to grams using its molar mass to find the theoretical yield in grams.
insert step 5: Calculate the percentage yield using the formula: \text{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced chemical equation. It allows chemists to determine the theoretical yield of a product from a given amount of reactants, which is essential for calculating percentage yield.
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Theoretical Yield

Theoretical yield is the maximum amount of product that can be produced from a given amount of reactants, assuming complete conversion and no losses. It is calculated using stoichiometric ratios derived from the balanced equation, providing a benchmark against which actual yield can be compared.
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Percentage Yield

Percentage yield is a measure of the efficiency of a reaction, calculated by dividing the actual yield by the theoretical yield and multiplying by 100. This value indicates how much of the expected product was actually obtained, helping to assess the effectiveness of the reaction and any potential losses.
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Related Practice
Textbook Question

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. If 5.00 g of sulfuric acid and 5.00 g of lead(II) acetate are mixed, calculate the number of grams of sulfuric acid and grams of acetic acid present in the mixture after the reaction is complete.

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Textbook Question

Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. If 5.00 g of sulfuric acid and 5.00 g of lead(II) acetate are mixed, calculate the number of grams of lead(II) acetate present in the mixture after the reaction is complete.

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Textbook Question

When benzene 1C6H62 reacts with bromine 1Br22, bromobenzene 1C6H5Br2 is obtained: C6H6 + Br2¡C6H5Br + HBr (a) When 30.0 g of benzene reacts with 65.0 g of bromine, what is the theoretical yield of bromobenzene?

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Textbook Question

When ethane 1C2H62 reacts with chlorine 1Cl22, the main product is C2H5Cl, but other products containing Cl, such as C2H4Cl2, are also obtained in small quantities. The formation of these other products reduces the yield of C2H5Cl. (a) Calculate the theoretical yield of C2H5Cl when 125 g of C2H6 reacts with 255 g of Cl2, assuming that C2H6 and Cl2 react only to form C2H2Cl and HCl. (b) Calculate the percent yield of C2H5Cl if the reaction produces 206 g of C2H5Cl.

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Textbook Question

Hydrogen sulfide is an impurity in natural gas that must be removed. One common removal method is called the Claus process, which relies on the reaction: 8 H2S1g2 + 4 O21g2¡S81l2 + 8 H2O1g2 Under optimal conditions the Claus process gives 98% yield of S8 from H2S. If you started with 30.0 g of H2S and 50.0 g of O2, how many grams of S8 would be produced, assuming 98% yield?

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Textbook Question

When hydrogen sulfide gas is bubbled into a solution of sodium hydroxide, the reaction forms sodium sulfide and water. How many grams of sodium sulfide are formed if 1.25 g of hydrogen sulfide is bubbled into a solution containing 2.00 g of sodium hydroxide, assuming that the sodium sulfide is made in 92.0% yield?

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