/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 In Example 4.3 you found that a ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In Example 4.3 you found that a mixture of \(\mathrm{CO}\) and \(\mathrm{H}_{2}\) produced \(407 \mathrm{g} \mathrm{CH}_{3} \mathrm{OH}\) $$ \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(\ell) $$ If only \(332 \mathrm{g}\) of \(\mathrm{CH}_{3} \mathrm{OH}\) is actually produced, what is the percent yield of the compound?

Short Answer

Expert verified
The percent yield is approximately 81.5%.

Step by step solution

01

Understand Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction. It is calculated using the formula:\[\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%\]where the actual yield is the amount of product obtained from the experiment, and the theoretical yield is the amount calculated based on stoichiometry.
02

Identify Given Values

From the problem, we know the actual yield of \( \mathrm{CH}_3\mathrm{OH} \) is \( 332 \text{ g} \). The theoretical yield is \( 407 \text{ g} \) as calculated in the earlier example.
03

Apply the Percent Yield Formula

Substitute the given values into the percent yield formula:\[\text{Percent Yield} = \left( \frac{332}{407} \right) \times 100\%\]
04

Calculate the Percent Yield

Perform the division and multiplication:\[\frac{332}{407} \approx 0.815\]Then,\[0.815 \times 100\% \approx 81.5\%\]So, the percent yield is approximately \( 81.5\% \).

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 91Ó°ÊÓ!

Key Concepts

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

Chemical Reaction Efficiency
Understanding the efficiency of a chemical reaction is essential when evaluating production processes. Chemical reaction efficiency is a measure of how well the reactants are converted into products. This efficiency can be influenced by various factors, such as:
  • Reaction conditions (temperature, pressure)
  • Purity of reactants
  • Presence of catalysts
  • Side reactions
A highly efficient reaction will result in a greater quantity of the desired product from the given reactants. This is crucial for economic and environmental considerations, as efficient processes minimize waste and use fewer resources. In this context, understanding percent yield becomes very important as a specific measure of this efficiency.
Percent Yield Formula
The percent yield formula provides a straightforward way to quantify the efficiency of a chemical reaction. It is calculated as:\[\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%\]Here, actual yield is the amount of product actually obtained from performing the reaction. On the other hand, the theoretical yield is the amount of product predicted by stoichiometric calculations, assuming complete conversion of reactants without any losses. Using this formula, one can determine how close the experimental results are to the ideal predictions, providing a numeric value that indicates the reaction's effectiveness. An actual yield lower than the theoretical yield signifies that some efficiency losses occurred during the process.
Stoichiometry
Stoichiometry is the study of the quantitative relationships between the quantities of reactants and products in a chemical reaction. It leverages the balanced chemical equations to make these calculations. In our given problem, stoichiometry was already used to determine the theoretical yield of methanol, \( \mathrm{CH}_3\mathrm{OH} \).Balancing chemical reactions helps to ensure that the same amount of each atom is present on both sides of the reaction. For example, the equation:\[ \mathrm{CO}(\mathrm{g}) + 2 \mathrm{H}_{2}(\mathrm{g}) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\ell) \]shows that 1 mole of carbon monoxide reacts with 2 moles of hydrogen to produce 1 mole of methanol. Stoichiometric calculations use these ratios, enabling us to determine the theoretical yield based on the amounts of reactants available.
Actual and Theoretical Yield
In chemical reactions, understanding both the actual yield and the theoretical yield is crucial for evaluating reaction performance. The **theoretical yield** is the calculated maximum amount of product that could be formed from a given set of reactants. It is based solely on stoichiometric calculations and does not consider any practical losses or inefficiencies in the reaction process. The **actual yield**, however, is the amount of product that is actually produced and measured in the laboratory. This value is often lower than the theoretical yield due to factors such as side reactions, incomplete reactions, or loss of product during handling. By comparing these two yields using the percent yield formula, one can assess how well the reaction performed and identify potential areas for improvement in efficiency.

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

Hexane \(\left(\mathrm{C}_{6} \mathrm{H}_{14}\right)\) burns in air \(\left(\mathrm{O}_{2}\right)\) to give \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) (a) Write a balanced equation for the reaction. (b) If \(215 \mathrm{g}\) of \(\mathrm{C}_{6} \mathrm{H}_{14}\) is mixed with \(215 \mathrm{g}\) of \(\mathrm{O}_{2},\) what masses of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) are produced in the reaction? (c) What mass of the excess reactant remains after the hexane has been burned?

Silicon and hydrogen form a series of compounds with the general formula \(\mathrm{Si}_{x} \mathrm{H}_{y}\). To find the formula of one of them, a 6.22 -g sample of the compound is burned in oxygen. All of the Si is converted to \(11.64 \mathrm{g}\) of \(\mathrm{SiO}_{2},\) and all of the \(\mathrm{H}\) is converted to \(6.980 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\). What is the empirical formula of the silicon compound?

Your body deals with excess nitrogen by excreting it in the form of urea, \(\mathrm{NH}_{2} \mathrm{CONH}_{2}\). The reaction producing it is the combination of arginine \(\left(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{N}_{4} \mathrm{O}_{2}\right)\) with water to give urea and ornithine \(\left(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{2}\right)\) \(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{N}_{4} \mathrm{O}_{2}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}+\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{N}_{2} \mathrm{O}_{2}\) Arginine Urea Ornithine If you excrete 95 mg of urea, what mass of arginine must have been used? What mass of ornithine must have been produced?

Mesitylene is a liquid hydrocarbon. Burning 0.115 g of the compound in oxygen gives \(0.379 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.1035 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula of mesitylene?

A major source of air pollution years ago was the metals industry. One common process involved "roasting" metal sulfides in the air: $$ 2 \mathrm{PbS}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{PbO}(\mathrm{s})+2 \mathrm{SO}_{2}(\mathrm{g}) $$ If you heat 2.5 mol of \(\mathrm{PbS}\) in the air, what amount of \(\mathrm{O}_{2}\) is required for complete reaction? What amounts of \(\mathrm{PbO}\) and \(\mathrm{SO}_{2}\) are expected?

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.