DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction: $$2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}+\mathrm{C}_{2} \mathrm{HOCl}_{3} \longrightarrow \mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}+\mathrm{H}_{2} \mathrm{O}$$ In a government lab, 1142 g of chlorobenzene is reacted with 485 g of chloral. a. What mass of DDT is formed, assuming \(100 \%\) yield? b. Which reactant is limiting? Which is in excess? c. What mass of the excess reactant is left over? d. If the actual yield of DDT is \(200.0 \mathrm{g},\) what is the percent yield?

Short Answer

Expert verified
a. The mass of DDT formed, assuming 100% yield, is \(1038.16~g\). b. The limiting reactant is chloral (\(C_2HOCl_3\)), and the reactant in excess is chlorobenzene (\(C_6H_5Cl\)). c. The mass of the excess reactant leftover is \(482.69~g\) of chlorobenzene. d. The percent yield of DDT in the reaction is \(19.28\%\).

Step by step solution

01

Calculating moles of reactants

First, we need to find the number of moles of each reactant using the respective molar mass. \(C_6H_5Cl\) has a molar mass of \((6 \cdot 12.01)+(5 \cdot 1.01)+(1 \cdot 35.45)=112.56~g/mol\). \(C_2HOCl_3\) has a molar mass of \((2 \cdot 12.01)+(1 \cdot 1.01)+(3 \cdot 35.45)=165.37~g/mol\). So for chlorobenzene, moles = \(\frac{1142}{112.56} = 10.15 ~mol\) For chloral, moles = \(\frac{485}{165.37} = 2.93 ~mol\)
02

Identifying the limiting reactant

To identify the limiting reactant, we first need to find the mole ratio of the reactants according to the balanced chemical equation: \(2~C_6H_5Cl + C_2HOCl_3 \rightarrow C_{14}H_9Cl_5 + H_2O\) The mole ratio of \(C_6H_5Cl\) to \(C_2HOCl_3\) is 2:1. Now, let's find the mole ratio using the given moles of reactants: \(\frac{10.15~mol}{2} = 5.075\) Since the mole ratio of the given reactants (5.075) is greater than the required mole ratio (2:1), this means that chloral (C2HOCl3) is the limiting reactant, while chlorobenzene (C6H5Cl) is in excess.
03

Calculating the mass of DDT formed

Given that the reaction has 100% yield, we can calculate the amount of DDT formed based on the limiting reactant. The molar mass of DDT is \((14 \cdot 12.01) + (9 \cdot 1.01) + (5 \cdot 35.45) = 354.49~g/mol\). According to the balanced chemical equation, 1 mole of chloral produces 1 mole of DDT. So, mass of DDT = moles of DDT x molar mass of DDT = \(2.93 ~mol \cdot 354.49 ~g/mol = 1038.16 ~g\)
04

Calculating the mass of leftover reactant

To calculate the mass of the excess reactant leftover after the reaction, we first need to find how much of the chlorobenzene actually reacted using the mole ratio: Required moles of chlorobenzene = 2 x moles of limiting reactant = 2 x 2.93 = 5.86 mol. Then, let's calculate the leftover moles of chlorobenzene and their mass: Leftover moles of chlorobenzene = Initial moles - Required moles = 10.15 - 5.86 = 4.29 mol So, the mass of leftover chlorobenzene = \(4.29 ~mol \cdot 112.56 ~g/mol = 482.69 ~g\)
05

Calculating the percent yield

Now that we have actual yield and theoretical yield, we can calculate the percentage yield as follows: Percent Yield = \(\frac{Actual~Yield}{Theoretical~Yield} \cdot 100\) Percent Yield = \(\frac{200.0}{1038.16} \cdot 100 = 19.28 \%.\) To summarize the answers to each part of the exercise: a. The mass of DDT formed, assuming 100% yield, is 1038.16 g. b. The limiting reactant is chloral (C2HOCl3), and the reactant in excess is chlorobenzene (C6H5Cl). c. The mass of the excess reactant leftover is 482.69 g of chlorobenzene. d. The percent yield of DDT in the reaction is 19.28%.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Limiting Reactant
Understanding the limiting reactant is crucial when working through stoichiometry problems. It's akin to a bottleneck in a production line—this is the reactant that runs out first, thus limiting the amount of product that can be formed in a chemical reaction. To determine the limiting reactant, calculate the number of moles of each reactant given their masses and molar masses. Then, use the balanced chemical equation to find the stoichiometric ratios and identify which reactant is used up first, given the actual amounts present.

To enhance this concept, remember that the limiting reactant dictates the theoretical maximum amount of product—once it's exhausted, the reaction can't proceed further, regardless of how much of the other reactants remains.
Theoretical Yield
The theoretical yield is the maximum amount of product that can be generated from a given amount of limiting reactant, assuming perfect conditions where no material is lost and the reaction goes to completion. Calculating the theoretical yield requires the mole-to-mole conversion between reactants and products, as dictated by the balanced chemical equation. Imagine it as the best-case scenario outcome.

Incorporating this into a real-world context, imagine you're baking cookies and you have the exact ingredients for 24 cookies—the theoretical yield is 24 cookies, assuming you followed the recipe perfectly and no dough was left sticking to the bowl or your fingers.
Percent Yield
The concept of percent yield comes into play when comparing the efficiency of a laboratory reaction versus the ideal scenario. It's calculated by dividing the actual yield (the amount of product actually obtained from the experiment) by the theoretical yield and then multiplying by 100 to get a percentage. The percent yield can tell us a lot about a reaction, like how practical it might be on a larger scale or if there were any errors or unexpected side reactions during the process.

For example, if you only obtained 18 cookies from the expected 24, your percent yield would be \(\frac{18}{24} \cdot 100 = 75\%\), reflecting the efficiency of your cookie-making process.
Molar Mass
The molar mass of a substance is the mass of one mole of its particles (atoms, molecules, ions, etc.) and is represented in grams per mole (g/mol). It is determined by summing the atomic masses of all atoms present in a molecule, as found on the periodic table. Molar mass serves as a conversion factor between the mass of a substance and the number of moles, and is fundamental to all stoichiometry calculations.

If you're working with a complex compound, like the insecticide DDT in our exercise, calculating its molar mass accurately is key for determining the stoichiometry of the reaction and thus, everything from the limiting reactant to the theoretical and percent yields.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Bornite \(\left(\mathrm{Cu}_{3} \mathrm{FeS}_{3}\right)\) is a copper ore used in the production of copper. When heated, the following reaction occurs: $$2 \mathrm{Cu}_{3} \mathrm{FeS}_{3}(s)+7 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{Cu}(s)+2 \mathrm{FeO}(s)+6 \mathrm{SO}_{2}(g)$$ If 2.50 metric tons of bornite is reacted with excess \(\mathrm{O}_{2}\) and the process has an \(86.3 \%\) yield of copper, what mass of copper is produced?

A 0.4230-g sample of impure sodium nitrate was heated, converting all the sodium nitrate to 0.2864 g of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

Boron consists of two isotopes, \(^{10} \mathbf{B}\) and \(^{11} \mathbf{B}\). Chlorine also has two isotopes, \(^{35} \mathrm{Cl}\) and \(^{37} \mathrm{Cl}\). Consider the mass spectrum of \(\mathrm{BCl}_{3}\) How many peaks would be present, and what approximate mass would each peak correspond to in the \(\mathrm{BCl}_{3}\) mass spectrum?

Gallium arsenide, GaAs, has gained widespread use in semiconductor devices that convert light and electrical signals in fiber-optic communications systems. Gallium consists of \(60 . \%^{69} \mathrm{Ga}\) and \(40 . \%^{71} \mathrm{Ga}\). Arsenic has only one naturally occurring isotope, \(^{75}\)As. Gallium arsenide is a polymeric material, but its mass spectrum shows fragments with the formulas GaAs and \(\mathrm{Ga}_{2} \mathrm{As}_{2}\). What would the distribution of peaks look like for these two fragments?

The reusable booster rockets of the U.S. space shuttle employ a mixture of aluminum and ammonium perchlorate for fuel. A possible equation for this reaction is $$\begin{aligned}3 \mathrm{Al}(s)+3 \mathrm{NH}_{4} \mathrm{ClO}_{4}(s) & \longrightarrow \\ \mathrm{Al}_{2} \mathrm{O}_{3}(s)+& \mathrm{AlCl}_{3}(s)+3 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\end{aligned}$$ What mass of \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) should be used in the fuel mixture for every kilogram of Al?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free