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Chapter 3: Problem 83

When benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) reacts with bromine \(\left(\mathrm{Br}_{2}\right)\), bromobenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\right)\) is obtained: $$ \mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Br}_{2} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}+\mathrm{HBr} $$ (a) When \(30.0 \mathrm{~g}\) of benzene reacts with \(65.0 \mathrm{~g}\) of bromine, what is the theoretical yield of bromobenzene? (b) If the actual yield of bromobenzene is \(42.3 \mathrm{~g}\), what is the percentage yield?

Short Answer

Expert verified
(a) The theoretical yield of bromobenzene is 63.9 g. (b) The percentage yield of bromobenzene is 66.2%.

Step by step solution

01

Determine the molar masses of benzene, bromine, and bromobenzene.

Calculate the molar mass of benzene (C6H6), bromine (Br2), and bromobenzene (C6H5Br) by adding up the molar masses of each element in the compounds according to the periodic table: Benzene (C6H6): \(6 \times 12.01 + 6 \times 1.01 = 72.06 \ \mathrm{g/mol}\) Bromine (Br2): \(2 \times 79.9 = 159.8 \ \mathrm{g/mol}\) Bromobenzene (C6H5Br): \(6 \times 12.01 + 5 \times 1.01 + 79.9 = 157.01 \ \mathrm{g/mol}\)
02

Find the number of moles of reactants.

Divide the given masses by their molar masses: Benzene: \(30.0 \ \mathrm{g} \div 72.06 \ \mathrm{g/mol} = 0.416 \ \mathrm{mol}\) Bromine: \(65.0 \ \mathrm{g} \div 159.8 \ \mathrm{g/mol} = 0.407 \ \mathrm{mol}\)
03

Find the limiting reactant.

Compare the molar ratio of the reactants to the coefficients in the balanced equation to determine which reactant is the limiting reactant: Benzene to Bromine: \(\frac{0.416}{1} \div \frac{0.407}{1} = 1.022\) Since the ratio is greater than the stoichiometric ratio (1:1), benzene is in excess and bromine is the limiting reactant.
04

Calculate the theoretical yield.

Use the moles of the limiting reactant (bromine) and the balanced chemical equation to find the moles of bromobenzene produced: \(0.407 \ \mathrm{mol \ Bromine} \times \frac{1 \ \mathrm{mol \ Bromobenzene}}{1 \ \mathrm{mol \ Bromine}} = 0.407 \ \mathrm{mol \ Bromobenzene}\) Now, convert the moles of bromobenzene to grams using its molar mass: Theoretical yield: \(0.407 \ \mathrm{mol} \times 157.01 \ \mathrm{g/mol} = 63.9 \ \mathrm{g}\) (a) The theoretical yield of bromobenzene is 63.9 g.
05

Determine the percentage yield.

Divide the actual yield by the theoretical yield and multiply by 100: Percentage yield: \(\frac{42.3 \ \mathrm{g}}{63.9 \ \mathrm{g}} \times 100 = 66.2 \% \) (b) The percentage yield of bromobenzene is 66.2%.

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Most popular questions from this chapter

(a) When a compound containing \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{O}\) is completely combusted in air, what reactant besides the hydrocarbon is involved in the reaction? (b) What products form in this reaction? (c) What is the sum of the coefficients in the balanced chemical equation for the combustion of acetone, \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}(l)\), in air?

(a) You are given a cube of silver metal that measures \(1.000\) \(\mathrm{cm}\) on each edge. The density of silver is \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\). How many atoms are in this cube? (b) Because atoms are spherical, they cannot occupy all of the space of the cube. The silver atoms pack in the solid in such a way that \(74 \%\) of the volume of the solid is actually filled with the silver atoms. Calculate the volume of a single silver atom. (c) Using the volume of a silver atom and the formula for the volume of a sphere, calculate the radius in angstroms of a silver atom.

A piece of aluminum foil \(1.00 \mathrm{~cm}^{2}\) and \(0.550-\mathrm{mm}\) thick is allowed to react with bromine to form aluminum bromide. (a) How many moles of aluminum were used? (The density of aluminum is \(2.699 \mathrm{~g} / \mathrm{cm}^{3}\).) (b) How many grams of aluminum bromide form, assuming the aluminum reacts completely?

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What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{HCO}_{2}\), molar mass \(=90.0 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\), molar mass \(=88 \mathrm{~g} / \mathrm{mol}\)
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