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

According to his prelaboratory theoretical yield calculations, a student's experiment should have produced \(1.44 \mathrm{~g}\) of magnesium oxide. When he weighed his product after reaction, only \(1.23 \mathrm{~g}\) of magnesium oxide was present. What is the student's percent yicld?

Short Answer

Expert verified
The student's percent yield of magnesium oxide in the experiment is 85.42%. This is calculated using the formula: Percent Yield = (Actual Yield / Theoretical Yield) × 100%, where Actual Yield is 1.23 g and Theoretical Yield is 1.44 g.

Step by step solution

01

Identify the given values

We are given the following values: - Theoretical Yield of magnesium oxide: 1.44 g - Actual Yield of magnesium oxide: 1.23 g
02

Apply the percent yield formula

Now, we will use the given values and plug them into the percent yield formula: Percent Yield = (Actual Yield / Theoretical Yield) × 100%
03

Calculate the percent yield

Substitute the given values and compute the result: Percent Yield = (1.23 g / 1.44 g) × 100%
04

Simplify the expression and find the answer

Calculate the division and multiply the result by 100%: Percent Yield = (0.8542) × 100% Percent Yield = 85.42% So, the student's percent yield of magnesium oxide in the experiment is 85.42%.

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.

Theoretical Yield
Theoretical yield is an important concept in chemistry. It refers to the maximum amount of product that can be produced in a chemical reaction. This prediction is based on the balanced equation and the initial quantities of reactants used. The theoretical yield is calculated by assuming that:
  • The reaction goes to completion.
  • All reactants are converted into the desired product without any side reactions.
  • No losses occur during the process.
In the context of the magnesium oxide experiment, the theoretical yield was calculated to be 1.44 grams. This means that, theoretically, if every atom of magnesium reacted with oxygen perfectly, the student should have obtained exactly 1.44 grams of magnesium oxide. However, achieving the theoretical yield in real-world lab settings is extremely rare due to various unavoidable factors.
Actual Yield
Actual yield refers to the amount of product that is actually produced when the chemical reaction is carried out. It often varies from the theoretical yield due to a range of factors, including:
  • Incomplete reactions where not all reactants form the desired product.
  • Side reactions that can produce different products.
  • Loss of product during processes like filtration or transfer.
In the scenario of the magnesium oxide experiment, the actual yield was 1.23 grams. This value represents what the student physically obtained after the reaction. Differences between actual and theoretical yields are typical in experiments and highlight the practical challenges in converting reactants to products efficiently.
Magnesium Oxide
Magnesium oxide (MgO) is a compound formed by the combination of magnesium and oxygen. It is commonly produced through a simple reaction where magnesium burns in the presence of oxygen:
\[ 2 \, Mg + O_{2} \rightarrow 2 \, MgO \]Magnesium oxide is known for its white, powdery form and high melting point. It is widely used across various industries:
  • As a refractory material in high-temperature furnaces.
  • In agriculture as a magnesium source for animals and plants.
  • In medicine, used to treat indigestion and as a magnesium supplement.
The study of this compound in experiments helps students understand chemical reactions and concepts like yield calculations, due to its straightforward reaction properties and easily measurable results.

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

For cach of the following balanced chemical cquations, calculate how many moles and how many grams of each product would be produced by the complete conversion of 0.50 mole of the reactant indicated in boldface. State clearly the mole ratio used for each conversion. a. \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\) b. \(\mathrm{CH}_{4}(g)+4 \mathrm{~S}(s) \rightarrow \mathrm{CS}_{2}(l)+2 \mathrm{H}_{2} \mathrm{~S}(g)\) c. \(\mathbf{P C l}_{3}(l)+3 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+3 \mathrm{HCl}(a q)\) d. \(\mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{NaHCO}_{3}(s)\)

The traditional method of analysis for the amount of chloride ion prescnt in a sample is to dissolve the sample in water and then slowly to add a solution of silver nitrate. Silver chloride is very insoluble in water, and by adding a slight excess of silver nitrate, it is possible to effectively remove all chloride ion from the sample. Suppose a \(1.054-g\) sample is known to contain \(10.3 \%\) chloride ion by mass. What mass of silver nitrate must be used to completely precipitate the chloride ion from the sample? What mass of silver chloride will be obtained?

Ammonium nitrate has been used as a high explosive because it is unstable and decomposes into several gaseous substances. The rapid expansion of the gaseous substances produces the explosive force. $$ \mathrm{NH}_{4} \mathrm{NO}_{3}(s) \rightarrow \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Calculate the mass of each product gas if \(1.25 \mathrm{~g}\) of ammonium nitrate reacts.

Small quantities of ammonia gas can be generated in the laboratory by heating an ammonium salt with a strong base. For example, ammonium chloride reacts with sodium hydroxide according to the following balanced equation: $$ \mathrm{NH}_{4} \mathrm{Cl}(s)+\mathrm{NaOH}(s) \rightarrow \mathrm{NH}_{3}(g)+\mathrm{NaCl}(s)+\mathrm{H}_{2} \mathrm{O}(g) $$ What mass of ammonia gas is produced if \(1.39 \mathrm{~g}\) of ammonium chloride reacts completely?

For cach of the following balanced reactions, calculate how many moles of each product would be produced by complete conversion of 0.50 mole of the reactant indicated in boldface. Indicate clearly the mole ratio used for the conversion. a. \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\) b. \(2 \mathrm{KClO}_{3}(s) \rightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) c. \(2 \mathrm{Al}(s)+6 \mathrm{HCl}(a q) \rightarrow 2 \mathrm{AlCl}_{3}(a q)+3 \mathrm{H}_{2}(g)\) d. \(\mathbf{C}_{3} \mathbf{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\)

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