Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Calculate the theoretical yield of the product (in moles) for each initial amount of reactants. $$\mathrm{Ti}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{TiCl}_{4}(s)$$ a. 4 \(\mathrm{mol} \mathrm{Ti}\) , 4 \(\mathrm{mol} \mathrm{Cl}_{2}\) b. \(7 \mathrm{mol} \mathrm{Ti}, 17 \mathrm{mol} \mathrm{Cl}_{2}\) c. \(12.4 \mathrm{mol} \mathrm{Ti}, 18.8 \mathrm{mol} \mathrm{Cl}_{2}\)

Short Answer

Expert verified
a. 4 moles of TiCl4, b. 7 moles of TiCl4, c. 9.4 moles of TiCl4

Step by step solution

01

Identify the Limiting Reactant for (a)

To calculate the theoretical yield for each set of reactants, determine the limiting reactant, which is the reactant that will be completely consumed first in the chemical reaction. Using the stoichiometry of the balanced equation, compare the mole ratio of reactants provided. For (a), we have 4 moles of Ti and 4 moles of Cl2 but the reaction requires 2 moles of Cl2 for every mole of Ti. Since we have exactly the amount of Cl2 required for 4 moles of Ti, neither is in excess. Thus, the reaction will run to completion without any limiting reactant, and the theoretical yield of TiCl4 will be equal to the moles of Ti, which is 4 moles.
02

Identify the Limiting Reactant for (b)

For (b), apply the same approach. With 7 moles of Ti and 17 moles of Cl2, calculate the number of moles of Cl2 needed to react with the given Ti. This is done by multiplying the moles of Ti by the stoichiometric coefficient ratio of Cl2 to Ti from the balanced equation (2:1). Thus, 7 moles of Ti would require 7 * 2 = 14 moles of Cl2. Since we have 17 moles of Cl2, which is more than enough to react with 7 moles of Ti, Cl2 is in excess, and Ti is the limiting reactant. The theoretical yield of TiCl4 is therefore the same as the moles of Ti, which is 7 moles.
03

Identify the Limiting Reactant for (c)

For (c), with 12.4 moles of Ti and 18.8 moles of Cl2, find the number of moles of Cl2 required for 12.4 moles of Ti by the balanced equation. It's 12.4 * 2 = 24.8 moles of Cl2. Since we have only 18.8 moles of Cl2, which is not enough to react with all 12.4 moles of Ti, Cl2 is the limiting reactant. To find the theoretical yield of TiCl4, determine the moles of Ti that can react with the available Cl2. Since 18.8 moles of Cl2 is the limiting reactant, we divide 18.8 moles of Cl2 by 2 (according to the stoichiometry) to get 9.4 moles. Therefore, the theoretical yield of TiCl4 is 9.4 moles.

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 Determination
Understanding the limiting reactant is a fundamental concept in chemistry, significant for predicting the amount of product formed in a chemical reaction. The limiting reactant is the substance that is completely consumed first in a chemical reaction, and it determines the maximum amount of product that can be formed.

The process to identify the limiting reactant involves a few systematic steps:
  • Begin by balancing the chemical equation to ensure the law of conservation of mass is satisfied.
  • Calculate the mole ratio of the reactants provided in the reaction.
  • Compare the mole ratios to the stoichiometric coefficients in the balanced equation.
  • The reactant that will produce the lesser amount of product, when its moles are divided by its stoichiometric coefficient, is the limiting reactant.
For example, in the reaction between titanium (Ti) and chlorine gas (Cl2) to form titanium tetrachloride (TiCl4), if we start with 7 moles of Ti and 17 moles of Cl2, we must first perform stoichiometric calculations to find out which reactant will run out first. As Ti requires twice as much Cl2 for a complete reaction (following the equation Ti + 2Cl2 → TiCl4), Ti turns out to be the limiting reactant because it only has enough Cl2 to react with each of its atoms. Consequently, the amount of Ti initially present dictates the maximum amount of TiCl4 that can be produced.
Stoichiometry
Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. It is derived from the Greek words 'stoicheion' (element) and 'metron' (measure), meaning the measurement of elements. Stoichiometry is the bridge that allows chemists to relate the mass of substances to the amount in moles and, ultimately, to the number of atoms, molecules, or ions in a given chemical reaction.

To apply stoichiometry in a problem, follow these steps:
  • Ensure that the chemical equation is balanced to reflect the conservation of mass.
  • Using the balanced equation, identify the mole ratio of the reactants and the products.
  • Use the mole ratio to convert between moles of one substance to moles of another.
For instance, if we have the balanced chemical equation Ti + 2Cl2 → TiCl4, and we know the number of moles of the limiting reactant (say, Ti), we can use stoichiometry to calculate the number of moles of TiCl4 that can be produced by multiplying the moles of Ti by the mole ratio from the equation. This is how we can calculate the theoretical yield from the given reactants.
Chemical Reaction Balancing
Balancing chemical equations is a necessary step in the study of chemical reactions, as it ensures that matter is not created or destroyed during the reaction, in alignment with the law of conservation of mass. It is important to balance chemical equations so we can correctly use them to predict the results of reactions.

To balance a chemical equation, follow these guidelines:
  • Write down the unbalanced equation with the known reactants and products.
  • Count the number of atoms of each element on both the reactant and product sides.
  • Adjust the coefficients (the numbers in front of the reactants and products) to make the number of atoms of each element equal on both sides of the equation.
  • It may be necessary to use trial and error, adjusting coefficients systematically until the equation is balanced.
For example, balancing the reaction between titanium and chlorine gas to form titanium tetrachloride requires placing a coefficient of '2' in front of Cl2 to ensure that there are four chlorine atoms on each side of the equation: Ti + 2Cl2 → TiCl4. Once balanced, this equation can now be used for stoichiometric calculations and determining the theoretical yield of the reaction.

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

Nitric acid (HNO3) is a component of acid rain that forms when gaseous nitrogen dioxide pollutant reacts with gaseous oxygen and liquid water to form aqueous nitric acid. Write a balanced chemical equation for this reaction. (Note: This is a simplified representation of this reaction.)

Consider the balanced equation: $$2 \mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 3 \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)$$ Complete the table showing the appropriate number of moles of reactants and products. If the number of moles of a reactant is provided, fill in the required amount of the other reactant, as well as the moles of each product that forms. If the number of moles of a product is provided, fill in the required amount of cach reactant to make that amount of product, as well as the amount of the other product that is made.

What is a combustion reaction? Why are they important? Give an example.

Consider the reaction: $$2 \mathrm{NO}(g)+5 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ A reaction mixture initially contains 5 moles of \(\mathrm{NO}\) and 10 moles of \(\mathrm{H}_{2} .\) Without doing any calculations, determine which set of amounts best represents the mixture after the reactants have reacted as completely as possible. Explain your reasoning. \(\begin{equation}\begin{array}{l}{\text { a. } 1 \mathrm{mol} \mathrm{NO}, 0 \mathrm{mol} \mathrm{H}_{2}, 4 \mathrm{mol} \mathrm{NH}_{3}, 4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} \\ {\text { b. } 0 \mathrm{mol} \mathrm{NO}, 1 \mathrm{mol} \mathrm{H}_{2}, 5 \mathrm{mol} \mathrm{NH}_{3}, 5 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}\end{array} \end{equation} \begin{equation}\begin{array}{l}{\text { c. } 3 \text { mol } \mathrm{NO}, 5 \mathrm{mol} \mathrm{H}_{2}, 2 \mathrm{mol} \mathrm{NH}_{3}, 2 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} \\ {\text { d. } 0 \text { mol } \mathrm{NO}, 0 \mathrm{mol} \mathrm{H}_{2}, 4 \mathrm{mol} \mathrm{NH}_{3}, 4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}\end{array} \end{equation}\)

For the reaction shown, calculate the theoretical yield of product (in grams) for each initial amount of reactants. $$2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AlCl}_{3}(s)$$ \begin{equation} \begin{array}{l}{\text { a. } 2.0 \mathrm{g} \mathrm{Al}, 2.0 \mathrm{g} \mathrm{Cl}_{2}} \\ {\text { b. } 7.5 \mathrm{g} \mathrm{Al}, 24.8 \mathrm{g} \mathrm{Cl}_{2}} \\ {\text { c. } 0.235 \mathrm{g} \mathrm{Al}, 1.15 \mathrm{g} \mathrm{Cl}_{2}}\end{array} \end{equation}

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