Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 116

Consider the reaction for the production of \(\mathrm{NO}_{2}\) from \(\mathrm{NO}\) : $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ (a) If \(84.8 \mathrm{~L}\) of \(\mathrm{O}_{2}(g)\), measured at \(35^{\circ} \mathrm{C}\) and \(632 \mathrm{~mm} \mathrm{Hg}\), is allowed to react with \(158.2 \mathrm{~g}\) of \(\mathrm{NO}\), find the limiting reagent. (b) If \(97.3 \mathrm{~L}\) of \(\mathrm{NO}_{2}\) forms, measured at \(35^{\circ} \mathrm{C}\) and \(632 \mathrm{~mm}\) \(\mathrm{Hg}\), what is the percent yield?

Short Answer

Expert verified
First, determine the limiting reagent by converting the given volumes and masses to moles. Then, calculate the theoretical yield of NO2 based on the limiting reagent. Finally, use the actual volume of NO2 formed to find the actual yield and calculate the percent yield using the formula (Actual yield / Theoretical yield) * 100%.

Step by step solution

01

Calculate moles of O2

Use the ideal gas law, PV = nRT, to calculate the moles of (O2). Use P = 632 mmHg (converted to atm), V = 84.8 L, R = 0.0821 (atm L/mol K), and T = 35°C (converted to Kelvin).
02

Convert mmHg to atm for O2

Convert the pressure of (O2) from mmHg to atm by using the conversion factor 1 atm = 760 mmHg.
03

Convert Celsius to Kelvin for O2

Convert the temperature from degrees Celsius to Kelvin by adding 273.15 to the Celsius temperature.
04

Calculate moles of NO

Calculate the moles of NO using its molar mass. The molar mass of NO is 30.01 g/mol. Use the given mass of NO, 158.2 g.
05

Determine limiting reagent

According to the balanced reaction, 2 moles of NO react with 1 mole of O2 to produce 2 moles of NO2. Compare the mole ratio of NO and O2 to find the limiting reagent.
06

Calculate moles of NO2 formed

Using the limiting reagent, calculate the theoretical yield of NO2 in moles.
07

Calculate moles of NO2 actually formed

Use the ideal gas law again to calculate the moles of NO2 actually formed from the given volume (97.3 L), temperature, and pressure.
08

Calculate percent yield

Percent yield = (Actual yield / Theoretical yield) * 100%. Use the results from Step 7 and Step 6 to find the percent yield.

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.

Start your free trial

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.

Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It is based on the principle that matter is neither created nor destroyed during a chemical reaction; thus, the mass and moles of reactants will equal the mass and moles of products.

Using stoichiometry, we can predict how much of each substance is needed and what will be formed in a reaction. This involves balancing chemical equations, converting between grams, liters, and moles, and understanding the molar ratios of the substances involved.
Ideal Gas Law
The ideal gas law is an equation of state for a hypothetical ideal gas. It is a good approximation of the behavior of many gases under a variety of conditions, though it has its limitations. The law is usually stated as PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature in Kelvin.

To apply the ideal gas law, it's crucial to use consistent units, such as atmosphere for pressure, liters for volume, and Kelvin for temperature. This law allows chemists to calculate the number of moles of a gas from its pressure, volume, and temperature—a key step in solving many stoichiometry problems.
Molar Mass
The molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). It can be thought of as the 'molecular weight' of a compound and is calculated by adding together the atomic masses of the elements that comprise the compound.

Knowing the molar mass is essential for converting between grams of a substance and moles, which is a fundamental step in stoichiometry. Without accurate molar mass values, calculations involving chemical reactions will be erroneous.
Theoretical Yield
The theoretical yield is the amount of product expected from a chemical reaction based on the amounts of reactants used, according to the stoichiometric calculations. It's calculated by using the balanced chemical equation and the limiting reagent—the reactant that will be entirely consumed first in the reaction, thus limiting the amount of product that can be formed.

In the ideal case where no product is lost in the process and all reactants convert to products without side reactions, the theoretical yield represents the maximum amount of product that could be obtained.
Actual Yield
In contrast to the theoretical yield, the actual yield is the amount of product actually obtained from a chemical reaction. This value is usually less than the theoretical yield due to factors such as incomplete reactions, side reactions, loss of product during transfer, or measurement imperfections.

Reporting the actual yield is crucial in both research and industrial applications to understand the efficiency of a chemical process, and it is a key figure in calculating the percent yield of a reaction.
Percent Yield Calculation
The percent yield is a measure of the efficiency of a chemical reaction, expressed as the ratio of the actual yield to the theoretical yield, multiplied by 100%. It's calculated using the formula:
\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100% \]
This value helps chemists understand how well a reaction proceeded and whether there may have been errors in the process or unexpected side reactions. Percent yield can indicate the purity of the product and the economic viability of a chemical synthesis in industrial settings.

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

Convert each measurement to atm. (a) 921 torr (b) \(4.8 \times 10^{4} \mathrm{~Pa}\) (c) \(87.5 \mathrm{psi}\) (d) \(34.22 \mathrm{in} . \mathrm{Hg}\)

The U.S. record for highest recorded barometric pressure is \(31.85\) in. Hg, set in 1989 in Northway, Alaska. Convert this pressure to: (a) millimeters of mercury (b) atmospheres (c) torr (d) kilopascals

What is standard temperature and pressure (STP)? What is the molar volume of a gas at STP?

Consider the reaction for the synthesis of nitric acid: $$ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) $$ (a) If \(12.8 \mathrm{~L}\) of \(\mathrm{NO}_{2}(g)\), measured at \(\mathrm{STP}\), is allowed to react with \(14.9 \mathrm{~g}\) of water, find the limiting reagent and the theoretical yield of \(\mathrm{HNO}_{3}\) in grams. (b) If \(14.8 \mathrm{~g}\) of \(\mathrm{HNO}_{3}\) forms, what is the percent yield?

How many moles of gas must be forced into a \(3.5\)-L ball to give it a gauge pressure of \(9.4\) psi at \(25^{\circ} \mathrm{C}\) ? The gauge pressure is relative to atmospheric pressure. Assume that atmospheric pressure is \(14.7\) psi so that the total pressure in the ball is \(24.1 \mathrm{psi}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation
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