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Ammonia gas can be prepared by the following reaction: \(\mathrm{CaO}(\mathrm{s})+2 \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CaCl}_{2}(\mathrm{s})\) If \(112 \mathrm{g}\) of \(\mathrm{CaO}\) and \(224 \mathrm{g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) are mixed, the theoretical yield of \(\mathrm{NH}_{3}\) is \(68.0 \mathrm{g}\) (Study Question 20 ). If only \(16.3 \mathrm{g}\) of \(\mathrm{NH}_{3}\) is actually obtained, what is its percent yield?

Short Answer

Expert verified
The percent yield is approximately 23.97%.

Step by step solution

01

Understand the Problem

The problem provides the masses of reactants and the theoretical yield of ammonia (NH₃). We need to find the percent yield using the actual yield of ammonia obtained.
02

Identify Given Information

From the problem, we know: the theoretical yield of NH₃ is 68.0 g, and the actual yield is 16.3 g.
03

Recall the Percent Yield Formula

The percent yield is calculated using the formula: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \% \]
04

Plug Values into the Formula

Substitute the known values into the percent yield formula: \[ \text{Percent Yield} = \left( \frac{16.3 \, \text{g}}{68.0 \, \text{g}} \right) \times 100 \% \]
05

Calculate the Percent Yield

Perform the calculation: \( \begin{align*} \text{Percent Yield} &= \left( \frac{16.3}{68.0} \right) \times 100 \ &= 23.97 \% \end{align*} \)

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Key Concepts

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

Ammonia Production
Ammonia is an essential compound widely used in various industries, primarily as a fertilizer and for manufacturing different chemicals. The production of ammonia typically involves combining nitrogen and hydrogen gases under high pressure and temperature, in a process known as the Haber-Bosch method. However, ammonia can also be produced from chemical reactions involving compounds like calcium oxide (CaO) and ammonium chloride (NH₄Cl), as in our original exercise.
In these reactions, specific conditions such as temperature and pressure are controlled to maximize the production of ammonia gas. Understanding these chemical processes is crucial for efficient ammonia production, which plays a vital role in agriculture and industry.
The reaction produces more than just ammonia gas; it also generates by-products like water vapor and calcium chloride, which are often managed or used in subsequent processes.
Theoretical Yield
In chemical reactions, the theoretical yield is the maximum amount of product that can be generated from a given amount of reactants. It is a crucial concept in chemistry because it provides a benchmark for what is theoretically possible under ideal conditions.
Calculating the theoretical yield involves understanding the mole ratios from the balanced chemical equation. In our exercise, the chemical reaction tells us the ideal amount of ammonia that would be produced if every molecule of reactant is used up perfectly, without any side reactions or losses.
However, real-world reactions rarely achieve the theoretical yield due to various inefficiencies, such as incomplete reactions, side reactions, or loss of material during the process. This is why the actual yield often falls short, making the theoretical yield a valuable goal but not always attainable.
Stoichiometry
Stoichiometry is the mathematical approach to understanding the relationships between reactants and products in a chemical reaction. It's essentially the bookkeeping of chemistry that ensures each component of a reaction is accounted for.
To work with stoichiometry effectively, it's important to use a balanced chemical equation, which reflects the conservation of mass and molecules. In the original exercise, we use stoichiometry to determine the theoretical yield of ammonia. This involves converting the mass of reactants (CaO and NH₄Cl) to moles, using their molar masses, and then applying the mole ratios from the balanced equation to find out how much product (NH₃) should form.
This method ensures that we understand how reactants transform into products systematically and can predict the outcomes of chemical reactions precisely.
Chemical Reactions
Chemical reactions are processes that involve the transformation of substances, where reactants are converted into products. They are fundamental to understanding the changes that occur in chemical systems.
  • A chemical reaction must be represented with a balanced equation, which clearly shows the conservation of atoms and charge.
  • In the reaction from our exercise, solid calcium oxide and ammonium chloride react to produce ammonia gas, water vapor, and solid calcium chloride.
  • Each component in the reaction has a specific role, which must be understood in terms of electron movement, bond breaking, and formation.
These reactions can be categorized by type, such as synthesis, decomposition, single-replacement, or double-replacement, providing insights into the nature of the transformations involved.
Understanding chemical reactions is essential for predicting how substances interact and for designing new compounds and materials in scientific and industrial applications.

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Most popular questions from this chapter

An unknown compound has the formula \(\mathrm{C}_{x} \mathrm{H}_{1} \mathrm{O}_{2}\). You burn \(0.1523 \mathrm{g}\) of the compound and isolate \(0.3718 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.1522 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\). What is the empirical formula of the compound? If the molar mass is \(72.1 \mathrm{g} / \mathrm{mol}\), what is the molecular formula? (See Exercise 4.9.)

A sample of limestone and other soil materials is heated, and the limestone decomposes to give calcium oxide and carbon dioxide. $$ \mathrm{CaCO}_{3}(\mathrm{s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$ A \(1.506-\mathrm{g}\) sample of limestone-containing material gives \(0.558 \mathrm{g}\) of \(\mathrm{CO}_{2},\) in addition to \(\mathrm{CaO},\) after being heated at a high temperature. What is the mass percent of \(\mathrm{CaCO}_{3}\) in the original sample?

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?

Balance the following equations: (a) Reaction to produce "superphosphate" fertilizer \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow \mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}(\mathrm{aq})+\mathrm{CaSO}_{4}(\mathrm{s})\) (b) Reaction to produce diborane, \(B_{2} H_{6}\) \(\mathrm{NaBH}_{4}(\mathrm{s})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g})+\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) (c) Reaction to produce tungsten metal from tungsten (VI) oxide \(\mathrm{WO}_{3}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{W}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (d) Decomposition of ammonium dichromate \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{s}) \longrightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})\)

Balance the following equations: (a) \(\mathrm{Cr}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{Cr} \mathrm{Cl}_{3}(\mathrm{s})\) (b) \(\mathrm{SiO}_{2}(\mathrm{s})+\mathrm{C}(\mathrm{s}) \longrightarrow \mathrm{Si}(\mathrm{s})+\mathrm{CO}(\mathrm{g})\) (c) \(\mathrm{Fe}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{g})\)

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