Chapter 7: Problem 74
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.
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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.
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.
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.
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.