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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 \(\mathrm{HCl}\). (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
(a) The theoretical yield of C2H5Cl is 232.2 g when 125 g of C2H6 reacts with 255 g of Cl2. (b) The percent yield of C2H5Cl is 88.75% when the reaction produces 206 g of C2H5Cl.

Step by step solution

01

Find the limiting reactant

To compute the theoretical yield, we first need to determine the limiting reactant. To find the limiting reactant, we need the balanced chemical equation and the masses of both reactants provided. The balanced chemical equation is: \[C_2H_6 + Cl_2 \rightarrow C_2H_5Cl + HCl\] The mass of ethane provided is 125 g and the mass of chlorine is 255 g.
02

Compute the number of moles for each reactant

To determine the limiting reactant, we will calculate the number of moles for each reactant: Number of moles = mass (g) / molar mass Molar masses: Ethane (C2H6): 2(12.01) + 6(1.01) = 30.07 g/mol Chlorine (Cl2): 2(35.45) = 70.90 g/mol No. of moles of ethane (nC2H6): nC2H6 = 125 g / 30.07 g/mol = 4.16 mol No. of moles of chlorine (nCl2): nCl2 = 255 g / 70.90 g/mol = 3.60 mol
03

Calculate the molar ratio and determine the limiting reactant

Now, we will compare the mole ratio of the reactants to the stoichiometric ratio from the balanced chemical equation: Mole ratio of ethane to chlorine: 4.16 mol / 3.60 mol = 1.16 Stoichiometric ratio from the balanced chemical equation is 1:1. Since the mole ratio is slightly greater than the stoichiometric ratio (1.16 > 1), the chlorine (Cl2) is the limiting reactant.
04

Calculate the theoretical yield of C2H5Cl

As chlorine (Cl2) is the limiting reactant, we will now use the number of moles of Cl2 to calculate the theoretical yield of ethyl chloride (C2H5Cl). The molar mass of C2H5Cl is 64.5 g/mol. The balanced chemical equation tells us that 1 mole of Cl2 yields 1 mole of C2H5Cl. Theoretical yield = moles of Cl2 × (1 mol C2H5Cl / 1 mol Cl2) × molar mass of C2H5Cl = 3.60 mol × 1 × 64.5 g/mol = 232.2 g
05

Calculate the percent yield of C2H5Cl

We are given that the actual yield of ethyl chloride (C2H5Cl) is 206 g. Now, we will calculate the percent yield using the formula: Percent yield = (Actual yield / Theoretical yield) × 100 = (206 g / 232.2 g) × 100 = 88.75% Hence, the percent yield of ethyl chloride (C2H5Cl) is 88.75%.

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Key Concepts

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

Limiting Reactant
Understanding the concept of the limiting reactant is essential in the realm of chemical reactions, as it determines the amount of product that can be formed. It refers to the reactant that will be completely used up first, limiting the reaction from proceeding further and dictating the theoretical yield.

In the provided exercise, the limiting reactant is found by comparing the amount of moles of each reactant with the stoichiometric ratio from the balanced chemical equation. The reactant that has fewer moles than required by the stoichiometric ratio is the one that limits the reaction, and it is essential to identify this because no more product can be created once this reactant is exhausted.
Mole Ratio
The mole ratio is a crucial component in stoichiometry, which is the ratio between the amounts in moles of any two compounds involved in a chemical reaction.

It is derived from the balanced equation and used to calculate how much of a reactant is needed or how much of a product will be formed. In the context of the exercise, the stoichiometric ratio is 1:1 for ethane to chlorine, which informs us about the proportionate relationship in which these reactants combine to form products.
Percent Yield
Percent yield is a measure of the efficiency of a chemical reaction, defined as the ratio of the actual yield to the theoretical yield, multiplied by 100.

A percent yield of 100% indicates that the reaction is perfectly efficient, while a lower percent yield suggests losses due to incomplete reactions, side reactions, or measurement errors. Practicality aside, these losses can help students understand where the discrepancies in their experiments might be occurring and encourage troubleshooting to achieve closer to the theoretical yield.
Stoichiometry

Stoichiometry in Chemical Equations

At its core, stoichiometry provides the quantitative relationship between reactants and products in a chemical reaction.

It involves calculations using balanced chemical equations to understand the amounts of substances consumed and produced. Stoichiometry is the foundation for theoretical yield calculations, as demonstrated in the example where it's used to predict the mass of ethyl chloride from given amounts of reactants.
Chemical Reactions
Chemical reactions are processes by which atoms or molecules are rearranged to form new substances.

Understanding chemical reactions allows us to predict products and their amounts. Through calculations like those in the exercise, students grasp practical concepts such as how much of a product can be expected from given reactants, being mindful of the reality that not all reactions proceed to completion, and some may produce multiple products or side products.

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Most popular questions from this chapter

Automotive air bags inflate when sodium azide, \(\mathrm{NaN}_{3}\), rapidly decomposes to its component elements: $$ 2 \mathrm{NaN}_{3}(s) \longrightarrow 2 \mathrm{Na}(s)+3 \mathrm{~N}_{2}(g) $$ (a) How many moles of \(\mathrm{N}_{2}\) are produced by the decomposition of \(1.50 \mathrm{~mol}\) of \(\mathrm{NaN}_{3}\) ? (b) How many grams of \(\mathrm{NaN}_{3}\) are required to form \(10.0 \mathrm{gg}\) of nitrogen gas? (c) How many grams of \(\mathrm{NaN}_{3}\) are required to produce \(10.0 \mathrm{ft}^{3}\) of nitrogen gas, about the size of an automotive air bag, if the gas has a density of \(1.25 \mathrm{~g} / \mathrm{L}\) ?

Without doing any detailed calculations (hut using a periodic table to give atomic weights), rank the following samples in order of increasing numbers of atoms: \(42 \mathrm{~g}\) of \(\mathrm{NaHCO}_{3}, 1.5 \mathrm{~mol} \mathrm{CO} 2,6.0 \times 10^{24} \mathrm{Ne}\) atoms.

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?

Aluminum hydroxide reacts with sulfuric acid as follows: Which is the limiting reactant when \(0.500 \mathrm{~mol} \mathrm{Al}(\mathrm{OH})_{3}\) and \(0.500 \mathrm{~mol} \mathrm{} \mathrm{H}_{2} \mathrm{SO}_{4}\) are allowed to react? How many moles of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) can form under these conditions? How many moles of the excess reactant remain after the completion of the reaction?

Balance the following equations: (a) \(\mathrm{Ca}_{3} \mathrm{P}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{PH}_{3}(g)\) (b) \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{AgNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow \mathrm{Ag}_{2} \mathrm{CO}_{3}(s)+\mathrm{NaNO}_{3}(a q)\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)\)

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