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

Nitrogen monoxide reacts with oxygen to give nitrogen dioxide. $$2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})$$ (a) You wish to react \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) in the correct stoichiometric ratio. The sample of NO has a volume of 150 mL. What volume of \(\mathrm{O}_{2}\) is required (at the same pressure and temperature)? (b) What volume of \(\mathrm{NO}_{2}\) (at the same pressure and temperature) is formed in this reaction?

Short Answer

Expert verified
75 mL of \(\mathrm{O}_{2}\) is required, and 150 mL of \(\mathrm{NO}_{2}\) is formed.

Step by step solution

01

Identify Reaction Stoichiometry

The balanced chemical equation is \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\), indicating that 2 volumes of NO react with 1 volume of \(\mathrm{O}_{2}\) to produce 2 volumes of \(\mathrm{NO}_{2}\). This implies a 2:1:2 stoichiometric ratio of \(\mathrm{NO}\):\(\mathrm{O}_{2}\):\(\mathrm{NO}_{2}\).
02

Calculate Volume of Oxygen Required

Given 150 mL of \(\mathrm{NO}\), and using the stoichiometric ratio 2:1, the required volume of \(\mathrm{O}_{2}\) is half the volume of \(\mathrm{NO}\). Thus, \( \frac{150 \text{ mL}}{2} = 75 \text{ mL} \) of \(\mathrm{O}_{2}\) is required.
03

Determine Volume of Nitrogen Dioxide Formed

Since the molar volume ratio of \(\mathrm{NO}\):\(\mathrm{NO}_{2}\) is 1:1, the volume of \(\mathrm{NO}_{2}\) formed is equal to the initial volume of \(\mathrm{NO}\) reacted. Therefore, 150 mL of \(\mathrm{NO}_{2}\) is formed.

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.

Balanced Chemical Equation
A balanced chemical equation is fundamental to understanding reactions. It ensures that the number of atoms for each element is equal on both sides of the equation. This balance reflects the conservation of mass principle, where no atoms are lost or gained during the reaction.
For the reaction: \[2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g})\]
  • The equation is balanced, meaning there are 2 nitrogen atoms and 4 oxygen atoms on both sides.
  • This balance helps us understand the reaction’s stoichiometry, which is crucial for accurately calculating reactant and product volumes.
Understanding a balanced equation provides the groundwork for solving stoichiometry problems in chemistry.
Volume Ratio
The volume ratio in a gas reaction provides insight into how much of each gas is used or produced. When gases react, their volumes are often directly proportional to their mole ratios if conditions are constant. This means we can use the coefficients from the balanced equation.
In our reaction:
  • The volume ratio of \(\mathrm{NO}\) to \(\mathrm{O}_2\) is 2:1.
  • The volume ratio of \(\mathrm{NO}\) to \(\mathrm{NO}_2\) is 1:1.
These ratios are vital when determining how much of a substance is needed or produced without needing to convert to moles.
Reaction Stoichiometry
Reaction stoichiometry involves using the balanced chemical equation to determine the proportions of reactants and products. It is crucial for calculating quantities needed in a chemical reaction.
The stoichiometric ratio is derived from the coefficients of each substance in the balanced equation. In our example:
  • 2 moles of \(\mathrm{NO}\) react with 1 mole of \(\mathrm{O}_2\) to produce 2 moles of \(\mathrm{NO}_2\).
  • This ratio helps in calculating the exact amounts needed or produced.
Applying stoichiometry, 150 mL of \(\mathrm{NO}\) requires 75 mL of \(\mathrm{O}_2\) and produces 150 mL of \(\mathrm{NO}_2\). This approach ensures precision in chemical reactions.
Gas Reactions
Gas reactions behave differently due to the properties of gases. One key feature is that, under the same conditions of temperature and pressure, equal volumes of gases contain equal numbers of molecules, according to Avogadro's law. This is why gas volume ratios can be directly taken from the coefficients in a balanced equation.
This implies that in the reaction:
  • 150 mL of \(\mathrm{NO}\) under the same conditions require 75 mL of \( \mathrm{O}_2 \).
  • The same 150 mL of \(\mathrm{NO}\) will produce 150 mL of \(\mathrm{NO}_2\).
Gas reactions are straightforward to analyze because volume, rather than mass, can be directly used. This simplifies calculations considerably, especially when dealing with reactions at constant temperature and pressure.

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

Nitrogen monoxide reacts with oxygen to give nitrogen dioxide: $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) $$ (a) Place the three gases in order of increasing rms speed at \(298 \mathrm{K}\) (b) If you mix \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) in the correct stoichiometric ratio and NO has a partial pressure of \(150 \mathrm{mm} \mathrm{Hg}\) what is the partial pressure of \(\mathrm{O}_{2} ?\) (c) After reaction between \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) is complete, what is the pressure of \(\mathrm{NO}_{2}\) if the NO originally had a pressure of \(150 \mathrm{mm} \mathrm{Hg}\) and \(\mathrm{O}_{2}\) was added in the correct stoichiometric amount?

You have two pressure-proof steel cylinders of equal volume, one containing \(1.0 \mathrm{kg}\) of \(\mathrm{CO}\) and the other containing \(1.0 \mathrm{kg}\) of acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\) (a) In which cylinder is the pressure greater at \(25^{\circ} \mathrm{C} ?\) (b) Which cylinder contains the greater number of molecules?

A Chlorine gas \(\left(\mathrm{Cl}_{2}\right)\) is used as a disinfectant in municipal water supplies, although chlorine dioxide \(\left(\mathrm{ClO}_{2}\right)\) and ozone are becoming more widely used. \(\mathrm{ClO}_{2}\) is a better choice than \(\mathrm{Cl}_{2}\) in this application because it leads to fewer chlorinated by-products, which are themselves pollutants. (a) How many valence electrons are in \(\mathrm{ClO}_{2} ?\) (b) The chlorite ion, \(\mathrm{ClO}_{2}^{-}\), is obtained by reducing ClO \(_{2} .\) Draw a possible electron dot structure for \(\mathrm{ClO}_{2}^{-} .\) (Cl is the central atom.) (c) What is the hybridization of the central Cl atom in \(\mathrm{ClO}_{2}^{-} ?\) What is the shape of the ion? (d) Which species has the larger bond angle, \(\mathbf{O}_{3}\) or \(\mathrm{ClO}_{2}^{-} ?\) Explain briefly. (e) Chlorine dioxide, \(\mathrm{ClO}_{2},\) a yellow-green gas, can be made by the reaction of chlorine with sodium chlorite: \(2 \mathrm{NaClO}_{2}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NaCl}(\mathrm{s})+2 \mathrm{ClO}_{2}(\mathrm{g})\) Assume you react \(15.6 \mathrm{g}\) of \(\mathrm{NaClO}_{2}\) with chlorine gas, which has a pressure of \(1050 \mathrm{mm}\) Hg in a \(1.45-\mathrm{L}\). flask at \(22^{\circ} \mathrm{C}\). What mass of \(\mathrm{ClO}_{2}\) can be produced?

Calculate the rms speed for CO molecules at \(25^{\circ} \mathrm{C}\) What is the ratio of this speed to that of Ar atoms at the same temperature?

What is the total pressure in atmospheres of a gas mixture that contains \(1.0 \mathrm{g}\) of \(\mathrm{H}_{2}\) and \(8.0 \mathrm{g}\) of Ar in a 3.0-L container at \(27^{\circ} \mathrm{C}\) ? What are the partial pressures of the two gases?

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