/*! 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 46 Consider the following chemical ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the following chemical equation. $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ If \(25.0 \mathrm{mL} \mathrm{NO}_{2}\) gas is completely converted to \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas under the same conditions, what volume will the \(\mathrm{N}_{2} \mathrm{O}_{4}\) occupy?

Short Answer

Expert verified
The Nâ‚‚Oâ‚„ gas will occupy a volume of 12.5 mL after the complete conversion of 25.0 mL NOâ‚‚ gas, based on the given balanced chemical equation and applying Avogadro's law.

Step by step solution

01

Understand the balanced equation

The given balanced chemical equation is: \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)\) This equation tells us that 2 moles of NOâ‚‚ react with each other to form 1 mole of Nâ‚‚Oâ‚„.
02

Identify the given information

We are given the following information: - Volume of NOâ‚‚ gas = 25.0 mL
03

Apply the relation between volumes of gases in a chemical reaction

According to Avogadro's law, equal volumes of all gases, at the same temperature and pressure, have the same number of molecules. We can apply this relation as follows for the given chemical reaction: Volume of Nâ‚‚Oâ‚„ gas / Volume of NOâ‚‚ gas = moles of Nâ‚‚Oâ‚„ / moles of NOâ‚‚ Since the balanced equation tells us that 2 moles of NOâ‚‚ react to produce 1 mole of Nâ‚‚Oâ‚„, we can write: Volume of Nâ‚‚Oâ‚„ gas / Volume of NOâ‚‚ gas = 1/2
04

Calculate the volume of Nâ‚‚Oâ‚„ gas

Now, we can rearrange the equation we found in Step 3 to solve for the volume of Nâ‚‚Oâ‚„ gas: Volume of Nâ‚‚Oâ‚„ gas = (1/2) * Volume of NOâ‚‚ gas Plugging in the given volume of NOâ‚‚ gas: Volume of Nâ‚‚Oâ‚„ gas = (1/2) * 25.0 mL Volume of Nâ‚‚Oâ‚„ gas = 12.5 mL Therefore, the Nâ‚‚Oâ‚„ gas will occupy a volume of 12.5 mL after the complete conversion of NOâ‚‚ gas.

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.

Understanding Chemical Equations
A chemical equation is a symbolic representation of a chemical reaction. It shows the reactants and products involved in the reaction along with the proportions in which they react. For example, the chemical equation \(2 \ \text{NO}_2(g) \longrightarrow \text{N}_2\text{O}_4(g)\) indicates that two molecules of nitrogen dioxide (\(\text{NO}_2\)) gas react to form one molecule of dinitrogen tetroxide (\(\text{N}_2\text{O}_4\)) gas.
This balanced equation is crucial as it maintains the Law of Conservation of Mass, meaning the quantity of each element is the same before and after the reaction.
When interpreting chemical equations, remember these points:
  • Coefficients (the numbers before the formulas) tell you the number of moles of each substance involved.
  • Physical states of reactants and products are indicated by letters in parentheses (like (g) for gas).
  • The arrow in the equation points from reactants to products, showing the direction of the reaction.
Using this knowledge helps in predicting the amounts of products formed and reactants required in a chemical reaction.
Exploring Avogadro's Law
Avogadro's Law is a fundamental principle in chemistry that relates the volume of a gas to the amount of substance present. It states that under the same conditions of temperature and pressure, equal volumes of different gases contain an equal number of molecules.
This law is invaluable when studying gaseous reactions, as it allows for a straightforward relationship between moles and volume. It can be expressed mathematically as:
  • \( V \propto n \) or \( \frac{V}{n} = k \)
where:
  • \(V\) is the volume of the gas.
  • \(n\) is the number of moles of gas.
  • \(k\) is a constant.
In the given exercise, Avogadro's law is applied to relate the volume of \(\text{NO}_2\) gas to \(\text{N}_2\text{O}_4\) based on their mole ratio from the balanced chemical equation. Since 2 moles of \(\text{NO}_2\) form 1 mole of \(\text{N}_2\text{O}_4\), the volume of \(\text{N}_2\text{O}_4\) is half that of \(\text{NO}_2\) gas, assuming the same conditions.
The Concept of Gas Volume Conversion
Gas volume conversion is essential to interpret the results of reactions involving gases. Often, we start off knowing the volume of a reactant gas and need to find the volume of a product gas, like in the given problem. This is made simpler by using the stoichiometric relationships provided by the balanced chemical equation and Avogadro's law.
Here's how you can approach gas volume conversions with ease:
  • Identify the volumes of gases given in the problem.
  • Determine the mole ratio from the balanced chemical equation (reactants to products).
  • Apply Avogadro's law to relate the initial and final volumes using the mole ratio.
In the exercise, we used these principles to find that the volume of \(\text{N}_2\text{O}_4\) was half of that of \(\text{NO}_2\) given the 2:1 mole ratio.
This conversion is an excellent example of how chemical equations and Avogadro's law come together to predict the behavior of gases in chemical reactions efficiently.

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

From the values in Table \(8-3\) for the van der Waals constant \(a\) for the gases \(\mathrm{H}_{2}, \mathrm{CO}_{2}, \mathrm{N}_{2},\) and \(\mathrm{CH}_{4},\) predict which of these gas molecules show the strongest intermolecular attractions.

Calculate the pressure exerted by \(0.5000\) mole of \(\mathrm{N}_{2}\) in a \(1.0000-\mathrm{L}\) container at \(25.0^{\circ} \mathrm{C}\) a. using the ideal gas law. b. using the van der Waals equation. c. Compare the results.

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of \(200 .\) L/min at \(1.50\) atm and ambient temperature. Air is added to the chamber at 1.00 atm and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g),\) three times as much oxygen as is necessary is reacted. Assuming air is \(21\) mole percent \(\mathrm{O}_{2}\) and \(79\) mole percent \(\mathrm{N}_{2}\), calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that \(95.0 \%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\). The remainder was present as carbon in \(\mathrm{CO}\). Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.

Sulfur trioxide, \(\mathrm{SO}_{3},\) is produced in enormous quantities each year for use in the synthesis of sulfuric acid. $$\begin{aligned}\mathrm{S}(s)+\mathrm{O}_{2}(g) & \longrightarrow \mathrm{SO}_{2}(g) \\\2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{SO}_{3}(g)\end{aligned}$$ What volume of \(\mathrm{O}_{2}(g)\) at \(350 .^{\circ} \mathrm{C}\) and a pressure of \(5.25\) atm is needed to completely convert \(5.00 \mathrm{g}\) sulfur to sulfur trioxide?

One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming \(\mathrm{Be}^{3+}\) ions) and that it gave an oxide with the formula \(\mathrm{Be}_{2} \mathrm{O}_{3}\). This resulted in a calculated atomic mass of \(13.5\) for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming \(\mathrm{Be}^{2+}\) ions) and that it gave an oxide with the formula BeO. This assumption gives an atomic mass of \(9.0 .\) In \(1894,\) A. Combes (Comptes Rendus 1894 p. 1221 ) reacted beryllium with the anion \(C_{5} \mathrm{H}_{7} \mathrm{O}_{2}^{-}\) and measured the density of the gaseous product. Combes's data for two different experiments are as follows:If beryllium is a divalent metal, the molecular formula of the product will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{2} ;\) if it is trivalent, the formula will be \(\mathrm{Be}\left(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2}\right)_{3} .\) Show how Combes's data help to confirm that beryllium is a divalent metal.

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.