Question

When ethane $\left({\mathrm{C}}_2{\mathrm{H}}_6\right)$ reacts with chlorine $\left({\mathrm{Cl}}_2\right),$ the main product is ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}$ , but other products containing $\mathrm{Cl},$ such as ${\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{Cl}}_2,$ are also obtained in small quantities. The formation of these other products reduces the yield of ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}$ (a) Calculate the theoretical yield of ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}$ when 125 $\mathrm{g}$ of ${\mathrm{C}}_2{\mathrm{H}}_6$ reacts with 255 $\mathrm{g}$ of ${\mathrm{Cl}}_2,$ assuming that ${\mathrm{C}}_2{\mathrm{H}}_6$ and ${\mathrm{Cl}}_2$ react only to form ${\mathrm{C}}_2{\mathrm{H}}_2\mathrm{Cl}$ and $\mathrm{HCl}.(\mathbf{b})$ Calculate the percent yield of ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}$ if the reaction produces 206 $\mathrm{g}$ of ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}$

Short answer

The theoretical yield of C2H5Cl is 232.3 g, and the percent yield of C2H5Cl is 88.7%.

Step-by-step solution

Write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is: $C_2{H}_6+C{l}_2\to C_2{H}_{5}Cl+HCl$

Calculate moles of reactants

Convert the mass of C2H6 and Cl2 to moles using their molar masses: Moles of C2H6 = $\frac{125g}{\mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{C}}_2{\mathrm{H}}_6}$ Moles of Cl2 = $\frac{255g}{\mathrm{Molar}\mathrm{mass}\mathrm{of}{\mathrm{Cl}}_2}$ Molar mass of C2H6 = 2×12.01 + 6×1.01 = 30.07 g/mol Molar mass of Cl2 = 2×35.45 = 70.90 g/mol Moles of C2H6 = $\frac{125g}{30.07g/mol}$ = 4.16 mol Moles of Cl2 = $\frac{255g}{70.90g/mol}$ = 3.60 mol

Determine the limiting reactant

To find the limiting reactant, compare the mole ratio of reactants with the stoichiometric ratio from the balanced equation: C2H6 : Cl2 (given ratio) = 4.16 : 3.60 C2H6 : Cl2 (stoichiometric ratio) = 1 : 1 Since the given ratio is higher for C2H6, Cl2 is the limiting reactant.

Calculate the theoretical yield of C2H5Cl

Use the molar ratio from the balanced equation to determine the maximum moles of C2H5Cl that can be formed from the limiting reactant: Moles of C2H5Cl = Moles of Cl2 = 3.60 mol Now, convert moles of C2H5Cl to grams: Theoretical yield of C2H5Cl = Moles of C2H5Cl × Molar mass of C2H5Cl Molar mass of C2H5Cl = 2×12.01 + 5×1.01 + 35.45 = 64.52 g/mol Theoretical yield of C2H5Cl = 3.60 mol × 64.52 g/mol = 232.3 g

Calculate the percent yield of C2H5Cl

With the actual mass of C2H5Cl produced being 206 g, we can now calculate the percent yield: Percent yield = $\frac{\mathrm{Actual}\mathrm{yield}}{\mathrm{Theoretical}\mathrm{yield}}\times 100$ Percent yield = $\frac{206g}{232.3g}\times 100$ = 88.7% Thus, the theoretical yield of C2H5Cl is 232.3 g, and the percent yield of C2H5Cl is 88.7%.

Key concepts

Chemical Reaction Stoichiometry

Understanding chemical reaction stoichiometry is fundamental to mastering chemistry. This concept involves determining the proportions of reactants and products involved in a chemical reaction. It is based on the law of conservation of mass, which states that matter is neither created nor destroyed during a chemical reaction.

Stoichiometry is employed when we use a balanced chemical equation to determine the amount of reactants necessary to form a desired quantity of product, or vice versa. For instance, in the balanced equation $C_2{H}_6+C{l}_2\to C_2{H}_{5}Cl+HCl$, the stoichiometric coefficients indicate a 1:1 mole ratio; for every mole of ethane ($C_2{H}_6$), one mole of chlorine ($C{l}_2$) is required.

When it comes to solving stoichiometric problems, here are key steps:

• Write a balanced chemical equation.
• Convert all given information, such as masses or volumes, to moles.
• Use the mole ratio from the balanced equation to calculate the number of moles of the desired substance.
• Convert moles back to the desired units, such as grams or liters.

Correctly implementing stoichiometry allows one to predict yields and optimize reactions in a laboratory or industrial setting, making it a powerful tool in the field of chemistry.

Limiting Reactant Determination

The concept of the limiting reactant is essential in predicting the amounts of products formed in a chemical reaction. It refers to the reactant that is completely consumed first in a chemical reaction, thus limiting the amount of product that can be formed. Understanding how to determine the limiting reactant ensures efficient use of resources in chemical processes.

To identify the limiting reactant, one must:

• Calculate the number of moles of each reactant.
• Compare the mole ratios of the reactants with the ratios provided by the balanced equation.
• Identify the reactant that will be consumed first based on the stoichiometry of the reaction.

In the provided exercise, we compared 4.16 moles of $C_2{H}_6$ with 3.60 moles of $C{l}_2$, and according to the balanced equation, they react in a 1:1 ratio. Since chlorine has fewer moles, it is the limiting reactant. Knowing which reactant will limit the reaction helps chemists to calculate the maximum possible yield—termed the theoretical yield—of the desired product.

Percent Yield Computation

Percent yield computation is a critical concept in both academic and industrial chemistry that measures the efficiency of a chemical reaction. The percent yield is the ratio of the actual yield (the amount of product actually obtained from a reaction) to the theoretical yield (the maximum amount of product expected based on stoichiometry) expressed as a percentage.

To calculate percent yield, we usually follow these steps:

• Determine the theoretical yield of the product from the limiting reactant using stoichiometry.
• Measure the actual yield of the product from the reaction.
• Use the formula $\text{Percent yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$.

In our exercise, the theoretical yield was found to be 232.3 grams of $C_2{H}_{5}Cl$ by stoichiometric calculations based on the limiting reactant chlorine. However, the actual yield reported was 206 grams. By applying the percent yield formula, we obtained an efficiency of 88.7%. This difference might be due to side reactions, incomplete reactions, or losses during product recovery. By analyzing percent yield, chemists can troubleshoot and refine reactions to minimize waste and improve efficiency.