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Chapter 11: Problem 15

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.

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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鈥檚 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.

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Most popular questions from this chapter

A 1.50 L constant volume calorimeter (Figure 5.12 ) contains \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g}) .\) The partial pressure of \(\mathrm{C}_{3} \mathrm{H}_{8}\) is 0.10 atm and the partial pressure of \(\mathrm{O}_{2}\) is 5.0 atm. The temperature is \(20.0^{\circ} \mathrm{C}\). A reaction occurs between the two compounds, forming \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\ell) .\) The heat from the reaction causes the temperature to rise to \(23.2^{\circ} \mathrm{C}\) (a) Write a balanced chemical equation for the reaction. (b) How many moles of \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})\) are present in the flask initially? (c) What is the mole fraction of \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})\) in the flask before reaction? (d) After the reaction, the flask contains excess oxygen and the products of the reaction, \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\ell) .\) What amount of unreacted \(\mathrm{O}_{2}(\mathrm{g})\) remains? (e) After the reaction, what is the partial pressure exerted by the \(\mathrm{CO}_{2}(\mathrm{g})\) in this system? (f) What is the partial pressure exerted by the excess oxygen remaining after the reaction?

A 1.007 -g sample of an unknown gas exerts a pressure of \(715 \mathrm{mm}\) Hg in a 452 -mL container at \(23^{\circ} \mathrm{C}\). What is the molar mass of the gas?

There are five compounds in the family of sulfur-fluorine compounds with the general formula \(\mathrm{S}_{x} \mathrm{F}_{y}\). One of these compounds is \(25.23 \%\) S. If you place \(0.0955 \mathrm{g}\) of the compound in a \(89-\mathrm{mL}\). flask at \(45^{\circ} \mathrm{C},\) the pressure of the gas is 83.8 mm Hg. What is the molecular formula of \(\mathrm{S}_{x} \mathrm{F}_{y} ?\)

A gaseous organofluorine compound has a density of \(0.355 \mathrm{g} / \mathrm{L}\) at \(17^{\circ} \mathrm{C}\) and \(189 \mathrm{mm}\) Hg. What is the molar mass of the compound?

A sample of nitrogen gas has a pressure of \(67.5 \mathrm{mm} \mathrm{Hg}\) in a 500 -mL. flask. What is the pressure of this gas sample when it is transferred to a 125 -m 1 . flask at the same temperature?
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