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

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

Step by step solution

01

Write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is: \(C_{2}H_{6} + Cl_{2} \rightarrow C_{2}H_{5}Cl + HCl\)
02

Calculate moles of reactants

Convert the mass of C2H6 and Cl2 to moles using their molar masses: Moles of C2H6 = \(\frac{125\,g}{\mathrm{Molar\,mass\,of\,C_{2}H_{6}}}\) Moles of Cl2 = \(\frac{255\,g}{\mathrm{Molar\,mass\,of\,Cl_{2}}}\) 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 = \(\frac{125\,g}{30.07\,g/mol}\) = 4.16 mol Moles of Cl2 = \(\frac{255\,g}{70.90\,g/mol}\) = 3.60 mol
03

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.
04

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
05

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 = \(\frac{\mathrm{Actual\,yield}}{\mathrm{Theoretical\,yield}} \times100\) Percent yield = \(\frac{206\,g}{232.3\,g} \times100\) = 88.7% Thus, the theoretical yield of C2H5Cl is 232.3 g, and the percent yield of C2H5Cl is 88.7%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 \( C_{2}H_{6} + Cl_{2} \rightarrow C_{2}H_{5}Cl + HCl \), the stoichiometric coefficients indicate a 1:1 mole ratio; for every mole of ethane (\( C_{2}H_{6} \)), one mole of chlorine (\( Cl_{2} \)) 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 \( Cl_{2} \), 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.

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Most popular questions from this chapter

Write "true" or "false" for each statement. (a) We balance chemical equations as we do because energy must be conserved. (b) If the reaction 2 \(\mathrm{O}_{3}(g) \rightarrow 3 \mathrm{O}_{2}(g)\) goes to completion and all \(\mathrm{O}_{3}\) is converted to \(\mathrm{O}_{2},\) then the mass of \(\mathrm{O}_{3}\) at the beginning of the reaction must be the same as the mass of \(\mathrm{O}_{2}\) at the end of the reaction. (c) You can balance the "water-splitting" reaction \(\mathrm{H}_{2} \mathrm{O}(I) \rightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) by writing it this way: \(\mathrm{H}_{2} \mathrm{O}_{2}(l) \rightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\)

Write the balanced chemical equations for (a) the complete combustion of acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right),\) the main ingredient in vinegar; (b) the decomposition of solid calcium hydroxide into solid calcium oxide (lime) and water vapor; (c) the combination reaction between nickel metal and chlorine gas.

Write balanced chemical equations to correspond to each of the following descriptions: (a) When sulfur trioxide gas reacts with water, a solution of sulfuric acid forms. (b) Boron sulfide, \(B_{2} S_{3}(s),\) reacts violently with water to form dissolved boric acid, \(H_{3} B O_{3},\) and hydrogen sulfide gas. (c) Phosphine, PH \(_{3}(g),\) combusts in oxygen gas to form water vapor and solid tetraphosphorus decaoxide. (d) When solid mercury(II) nitrate is heated, it decomposes to form solid mercury(II) oxide, gaseous nitrogen dioxide, and oxygen.(e) Copper metal reacts with hot concentrated sulfuric acid solution to form aqueous copper(II) sulfate, sulfur dioxide gas, and water.

The molecular formula of aspartame, the artificial sweetener marketed as NutraSweet, is \(\mathrm{C}_{14} \mathrm{H}_{18} \mathrm{N}_{2} \mathrm{O}_{5}\) . (a) What is the molar mass of aspartame? (b) How many moles of aspartame are present in 1.00 \(\mathrm{mg}\) of aspartame? (c) How many molecules of aspartame are present in 1.00 \(\mathrm{mg}\) of aspartame? (d) How many hydrogen atoms are present in 1.00 \(\mathrm{mg}\) of aspartame?

(a) One molecule of the antibiotic penicillin G has a mass of \(5.342 \times 10^{-21} \mathrm{g} .\) What is the molar mass of penicillin G? (b) Hemoglobin, the oxygen-carrying protein in red blood cells, has four iron atoms per molecule and contains 0.340\(\%\) iron by mass. Calculate the molar mass of hemoglobin.

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