How many molecules of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) can be prepared from \(15 \mathrm{C}_{2} \mathrm{H}_{4}\) molecules and \(8 \mathrm{Cl}_{2}\) molecules?

Short Answer

Expert verified
8 molecules of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) can be prepared.

Step by step solution

01

Establish the chemical reaction

The chemical reaction for the synthesis of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) from \(\mathrm{C}_{2} \mathrm{H}_{4}\) and \(\mathrm{Cl}_{2}\) is as follows: \[\mathrm{C}_{2} \mathrm{H}_{4} + \mathrm{Cl}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\]. This shows that one molecule of \(\mathrm{C}_{2} \mathrm{H}_{4}\) reacts with one molecule of \(\mathrm{Cl}_{2}\) to form one molecule of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\).
02

Identify the limiting reactant

With 15 molecules of \(\mathrm{C}_{2} \mathrm{H}_{4}\) and 8 molecules of \(\mathrm{Cl}_{2}\), \(\mathrm{Cl}_{2}\) is the limiting reactant because there are fewer molecules of \(\mathrm{Cl}_{2}\) available compared to \(\mathrm{C}_{2} \mathrm{H}_{4}\).
03

Calculate the number of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) molecules formed

Since \(\mathrm{Cl}_{2}\) is the limiting reactant and there are 8 molecules of it available, it will determine the maximum number of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) molecules that can be formed. Therefore, 8 molecules of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) can be formed.

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.

Chemical Reaction Equation
In chemistry, a chemical reaction equation represents the transformation of reactants into products. It is crucial in understanding how substances interact and what they produce when they do. For instance, in the exercise given, the equation is \[\mathrm{C}_{2} \mathrm{H}_{4} + \mathrm{Cl}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\. \] This simplifies the complex process of chemical reactions into something that can be studied and analyzed.

Every symbol and subscript in a chemical equation has a specific meaning; they indicate the elements involved, the number of atoms of each element, and how they are paired up during the reaction. Correct usage of these symbols allows us to predict the outcome of the reaction, such as the amount of product that can be formed from a given amount of reactants. In educational settings, mastering chemical equations is foundational, because they are the building blocks for more advanced topics in chemistry.
Stoichiometry
Stoichiometry is often described as the currency exchange of chemistry; it's about understanding the ratios and exchanges between different substances in a chemical reaction. The principle of stoichiometry is grounded in the law of conservation of mass, which states that in a chemical reaction, matter is neither created nor destroyed.

For the given exercise, stoichiometry is applied by looking at the mole-ratio between reactants and products. According to the equation, one molecule of \(\mathrm{C}_{2} \mathrm{H}_{4}\) reacts with one molecule of \(\mathrm{Cl}_{2}\) to form one molecule of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\). This 1:1:1 ratio is pivotal for determining how much product can be formed from given quantities of reactants. By using this ratio, if you had 15 molecules of \(\mathrm{C}_{2} \mathrm{H}_{4}\) and 8 molecules of \(\mathrm{Cl}_{2}\), you can easily predict the outcome, which brings us to the concept of the limiting reactant.
Molecular Synthesis
Central to chemistry is the concept of molecular synthesis, which is the process of building complex molecules from simpler ones. In our exercise, molecular synthesis involves creating \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) molecules from \(\mathrm{C}_{2} \mathrm{H}_{4}\) and \(\mathrm{Cl}_{2}\) molecules.

The success of this synthesis relies on having the correct amounts of reactants. If one reactant is present in a lesser quantity than required by the stoichiometric ratio, it limits the extent of the reaction — hence the term 'limiting reactant'. When the limiting reactant is fully consumed, the chemical reaction stops, regardless of the quantities of other reactants present. In the given exercise, \(\mathrm{Cl}_{2}\) is the limiting reactant. This means that no matter how much \(\mathrm{C}_{2} \mathrm{H}_{4}\) there is, the creation (synthesis) of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) molecules will cease once all \(\mathrm{Cl}_{2}\) molecules have been used in the reaction, resulting in the synthesis of 8 \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) molecules.

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

Complete and balance the following oxidation-reduction reactions, which give the highest possible oxidation state for the oxidized atoms. (a) \(\mathrm{K}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow\) (b) \(\mathrm{Ba}(s)+\mathrm{HBr}(a q) \rightarrow\) (c) \(\operatorname{Sn}(s)+I_{2}(s) \rightarrow\)

In a common experiment in the general chemistry laboratory, magnesium metal is heated in air to produce \(\mathrm{MgO} . \mathrm{MgO}\) is a white solid, but in these experiments it often looks gray, due to small amounts of \(\mathrm{Mg}_{3} \mathrm{N}_{2}, \mathrm{a}\) compound formed as some of the magnesium reacts with nitrogen. Write a balanced equation for each reaction.

Write a balanced molecular equation describing each of the following chemical reactions. (a) Solid calcium carbonate is heated and decomposes to solid calcium oxide and carbon dioxide gas. (b) Gaseous butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\), reacts with diatomic oxygen gas to yield gaseous carbon dioxide and water vapor. (c) Aqueous solutions of magnesium chloride and sodium hydroxide react to produce solid magnesium hydroxide and aqueous sodium chloride. (d) Water vapor reacts with sodium metal to produce solid sodium hydroxide and hydrogen gas.

The military has experimented with lasers that produce very intense light when fluorine combines explosively with hydrogen. What is the balanced equation for this reaction?

What volume of 0.0105-M HBr solution is required to titrate 125 mL of a 0.0100-M Ca(OH)_solution? \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{HBr}(a q) \rightarrow \mathrm{CaBr}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

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