Question

The reaction of ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) with chlorine gas produces \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) as its main product (along with HCl). In addition, the reaction invariably produces a variety of other minor products, including \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3},\) and others. Naturally, the production of these minor products reduces the yield of the main product. Calculate the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) if the reaction of \(300 .\) g of ethane with \(650 .\) g of chlorine produced \(490 .\) g of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}.\)

Short answer

The percent yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) in the reaction between ethane gas and chlorine gas is approximately 82.8%.

Step-by-step solution

Write the balanced equation

\ The balanced equation for the reaction between \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{Cl}_{2}\) is: \[ \mathrm{C}_{2} \mathrm{H}_{6} + \mathrm{Cl}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl} + \mathrm{HCl} \]

Calculate the moles of reactants

\ Given the masses of the reactants, we can calculate the moles of each reactant by dividing their masses by their respective molar masses. The molar masses of \(\mathrm{C}_{2} \mathrm{H}_{6}\) and \(\mathrm{Cl}_{2}\) are 30.07 g/mol and 70.90 g/mol, respectively. Moles of \(\mathrm{C}_{2} \mathrm{H}_{6} = \frac{300\:g}{30.07\:g/mol} = 9.98\:mol\) Moles of \(\mathrm{Cl}_{2} = \frac{650\:g}{70.90\:g/mol} = 9.17\:mol\)

Determine the limiting reactant

\ To find the limiting reactant, we need to compare the molar ratios of the reactants. The balanced equation shows that 1 mole of \(\mathrm{C}_{2} \mathrm{H}_{6}\) reacts with 1 mole of \(\mathrm{Cl}_{2}\). Divide the moles of each reactant by their respective stoichiometric coefficients: \[ \frac{9.98\:\text{mol } C_{2}H_{6}}{1} \quad \text{and} \quad \frac{9.17\:\text{mol } Cl_{2}}{1} \] Since 9.17 < 9.98, \(\mathrm{Cl}_{2}\) is the limiting reactant.

Calculate the theoretical yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\)

\ Using the moles of the limiting reactant (\(\mathrm{Cl}_2\)), calculate the moles of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) that can be produced by the stoichiometry of the balanced reaction: Moles of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl} = \mathrm{9.17\:mol} \, \frac{\mathrm{1\, mol\:C}_{2} \mathrm{H}_{5} \mathrm{Cl}} {\mathrm{1\: mol\: Cl}_{2}} = \mathrm{9.17\:mol}\) Now that we have the moles of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\), we can find the theoretical yield in grams, using the molar mass of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) (64.53 g/mol): Theoretical yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl} = 9.17\:mol \times 64.53\:g/mol = 591.79\:g\)

Calculate the percent yield

\ Now that we have the actual yield (490 g) and the theoretical yield (591.79 g), we can calculate the percent yield as follows: \[ \text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\%\] \[ \text{Percent yield} = \frac{490\:g}{591.79\:g} \times 100\% \approx 82.8\% \] The percent yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) in this reaction is approximately 82.8%.

Key concepts

Chemical Reaction Equations

Understanding chemical reaction equations is pivotal for students who are delving into the field of chemistry. These equations provide a symbolic representation of what occurs during a chemical reaction. For example, the balanced chemical equation \[ \text{C}_2\text{H}_6 + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{HCl} \] illustrates the reaction between ethane (\text{C}_2\text{H}_6) and chlorine gas (\text{Cl}_2), which yields ethyl chloride (\text{C}_2\text{H}_5\text{Cl}) and hydrogen chloride (HCl) as products. Each molecule in the equation is represented by its chemical formula, and the coefficients indicate the number of moles required to balance the reaction.

In educational settings, it's crucial to stress the importance of balancing chemical equations. The Law of Conservation of Mass states that mass cannot be created or destroyed in a chemical reaction, and therefore each element must have the same number of atoms on both the reactant and product sides of the equation. Balancing equations ensures that this law is respected. It's a foundational skill in chemistry that precedes more advanced concepts such as stoichiometry and yield calculations.

Limiting Reactant Determination

Determining the limiting reactant in a chemical reaction is essential when predicting the amount of product that will form. The limiting reactant, by definition, is the reactant that will be entirely consumed first, limiting the extent of the reaction, and thus, the amount of product that can be formed. During a chemical reaction, if one reactant is used up before the others, any further reaction will cease, even if other reactants are still available.

The process to identify the limiting reactant involves calculating the moles of each reactant and using stoichiometry—based on the coefficients in the balanced equation—to compare the moles. For instance,\[ \frac{9.17\text{ mol Cl}_2}{1} < \frac{9.98\text{ mol C}_2\text{H}_6}{1} \]This comparison indicates that chlorine gas (\text{Cl}_2) is the limiting reactant since it has fewer moles than ethane (\text{C}_2\text{H}_6) in the context of their reaction stoichiometry. To assist students in grasping this concept, it is beneficial to provide visual aids, such as pie charts or bar graphs, representing the moles of reactants and highlighting the one that is limiting.

Stoichiometry

Stoichiometry is the mathematical relationship between the quantities of reactants and products in a chemical reaction. It is derived from the balanced chemical equation, which provides the ratio of moles that react with each other and form products. For example, the reaction\[ \text{C}_2\text{H}_6 + \text{Cl}_2 \rightarrow \text{C}_2\text{H}_5\text{Cl} + \text{HCl} \]indicates that one mole of ethane reacts with one mole of chlorine gas to produce one mole of ethyl chloride and one mole of hydrogen chloride. The concept of stoichiometry can be daunting for some students, but it can be made simpler by using step-by-step processes to calculate the relationships.To further elucidate stoichiometry, it's often helpful to correlate it with everyday concepts such as recipes in cooking—just as a recipe dictates the exact amount of each ingredient needed to make a certain number of servings, a chemical equation outlines the exact amounts of reactants needed to get a certain amount of product. Understanding stoichiometry allows students to predict how much product can be made from given amounts of reactants (theoretical yield) and to calculate the efficiency of a reaction using percent yield. The latter is determined by comparing the actual yield from an experiment to the theoretical yield calculated from stoichiometric ratios.