Problem 83 - Chemistry The Central Science

Question

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 g of benzene reacts with 65.0 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?

Step-by-Step Solution

Answer

(a) The theoretical yield of bromobenzene is \(60.3 \: \text{g}\). (b) The percentage yield of bromobenzene is \(70.1\%\).

Step 1: Calculate the moles of reactants

First, we need to determine the moles of benzene and bromine used in the reaction. The molecular weight of benzene is 12.01 * 6 + 1.01 * 6 = 78.12 g/mol, while the molecular weight of bromine is 79.90 * 2 = 159.80 g/mol. Now, we can calculate the moles of each reactant: For benzene: Moles = mass (g) / molecular weight (g/mol) moles of benzene = 30 g / 78.12 g/mol = 0.384 moles For bromine: Moles = mass (g) / molecular weight (g/mol) moles of bromine = 65.0 g / 159.8 g/mol = 0.407 moles

Step 2: Determine the limiting reactant

To determine the limiting reactant, we need to find the mole ratio between benzene and bromine from the balanced equation: C6H6 + Br2 → C6H5Br + HBr The mole ratio is 1:1 for benzene to bromine. To find the limiting reactant, we need to divide the moles of each reactant by their respective coefficients in the balanced equation: For benzene: 0.384 moles / 1 = 0.384 For bromine: 0.407 moles / 1 = 0.407 Since 0.384 < 0.407, benzene is the limiting reactant.

Step 3: Calculate the theoretical yield of bromobenzene

In order to find the theoretical yield, we need to calculate the moles of bromobenzene formed based on the moles of the limiting reactant, benzene: moles of bromobenzene = moles of benzene * (1 mole bromobenzene / 1 mole benzene) moles of bromobenzene = 0.384 moles * (1/1) = 0.384 moles Now, we convert moles of bromobenzene to grams using its molecular weight (C=12.01, H=1.01, Br=79.90): theoretical_yield_bromobenzene (g) = moles * molecular weight (g/mol) theoretical_yield_bromobenzene (g) = 0.384 moles * (12.01 * 6 + 1.01 * 5 + 79.90 * 1) = 0.384 moles * 157.03 g/mol = 60.3 g So, the theoretical yield of bromobenzene is 60.3 g.

Step 4: Calculate the percentage yield of bromobenzene

Now, we can calculate the percentage yield using the actual yield mentioned in the problem (42.3 g) and the theoretical yield calculated in the previous step (60.3 g): percentage_yield = (actual_yield / theoretical_yield) * 100 percentage_yield = (42.3 g / 60.3 g) * 100 = 70.1 % The percentage yield of bromobenzene in this reaction is 70.1%.

Key Concepts

Limiting Reactant

In chemical reactions, reactants are combined in specific proportions dictated by their stoichiometry. But what happens when one reactant runs out before the others? That's when the concept of the limiting reactant, or limiting reagent, becomes crucial. This is the reactant that is entirely consumed first, limiting the amount of product that can be formed. It's like running out of bread when making sandwiches; no matter how much ham you have, you can't make more sandwiches without more bread.

In determining the limiting reactant, we need to use the stoichiometry of the reaction. By calculating the moles of each reactant and comparing them to the mole ratio required by the balanced chemical equation, we can identify which reactant will run out first. In the example given, we compared the mole amounts of benzene and bromine and found that benzene, with fewer moles, was the limiting reactant. This identification is a critical step in predicting the amounts of products in a reaction.

Mole Ratio

Understanding the mole ratio in a chemical reaction is akin to following a recipe while cooking - it tells us the proportions of ingredients we should mix to get our desired product. Mole ratio stems from the coefficients of each substance in a balanced chemical equation, which indicate the relative number of moles that react or form.

Illustrating with our benzene reaction, the balanced equation tells us that one mole of benzene reacts with one mole of bromine to produce one mole of bromobenzene and one mole of hydrogen bromide. This one-to-one-to-one ratio is essential in stoichiometric calculations, for it allows us to convert between moles of different substances. In solving our textbook problem, the mole ratio let us connect the moles of benzene, the limiting reactant, directly to the moles of bromobenzene, the product we were interested in, assuming all benzene molecules reacted.

Percentage Yield

After performing a chemical reaction, one of the most telling indicators of its efficiency is the percentage yield. This is a measure of the amount of product actually obtained (the actual yield) compared to the amount that ideally could be produced (the theoretical yield) according to stoichiometry calculations.

The formula for calculating percentage yield is quite straightforward: \[\begin{equation}percentage\ yield = \left(\frac{actual\ yield}{theoretical\ yield}\right) \times 100\end{equation}\] It's similar to a grade on a test; if you score 70 out of 100 points, your percentage score is 70%. In our example, the actual yield of bromobenzene was 42.3 grams, but theoretically, 60.3 grams could have been made if benzene was converted entirely into product with no losses or side reactions. Dividing the actual yield by the theoretical and multiplying by 100 we found that the reaction's efficiency was 70.1%. This metric is invaluable for chemists to gauge reaction success and for industries to assess the cost-efficiency of their production processes.