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Chapter 5: Problem 46

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 volume of \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas after the complete conversion of \(25.0 \mathrm{~mL}\) NO鈧� gas under the same conditions is \(12.5 \mathrm{~mL}\).

Step by step solution

01

Analyze the balanced chemical equation

The given chemical equation is: \[ 2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) \] From the equation, we can see that 2 moles of NO鈧� gas are needed to produce 1 mole of N鈧侽鈧� gas.
02

Determine the mole-to-mole ratio

The mole-to-mole ratio of NO鈧� to N鈧侽鈧� is given by the coefficients in the balanced chemical equation: \[\frac{\text{moles of N}_{2}\mathrm{O}_{4}}{\text{moles of NO}_{2}} = \frac{1}{2}\]
03

Use the mole-to-mole ratio to determine the volume of N鈧侽鈧�

Under the same conditions, the volume ratio of gases is equal to the mole-to-mole ratio, according to Avogadro's law. Therefore, we can write the volume ratio as: \[\frac{\text{volume of N}_{2}\mathrm{O}_{4}}{\text{volume of NO}_{2}} = \frac{\text{moles of N}_{2}\mathrm{O}_{4}}{\text{moles of NO}_{2}} = \frac{1}{2}\] Now, plug in the given volume of NO鈧� gas (25.0 mL): \[\frac{\text{volume of N}_{2}\mathrm{O}_{4}}{25.0\ \mathrm{mL}} = \frac{1}{2}\]
04

Solve for the volume of N鈧侽鈧�

Now, we need to solve for the volume of N鈧侽鈧� gas. We can do this by cross-multiplying and simplifying the equation obtained in Step 3: \[\text{volume of N}_{2}\mathrm{O}_{4} = 25.0\ \mathrm{mL} \times \frac{1}{2}\] \[\text{volume of N}_{2}\mathrm{O}_{4} = 12.5\ \mathrm{mL}\] So, the volume of N鈧侽鈧� gas after the complete conversion of 25.0 mL NO鈧� gas under the same conditions is 12.5 mL.

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Key Concepts

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

Avogadro's Law
Understanding Avogadro's law is crucial for solving problems involving gas volumes in chemistry. This principle states that equal volumes of all gases, at the same temperature and pressure, contain an equal number of molecules. In simpler terms, if you have two different gases under the same conditions of temperature and pressure, the volume of these gases will directly relate to the number of moles of the gas present.

Applying Avogadro's law in chemical equations allows you to predict how volumes of gaseous reactants and products will relate to each other, as long as they are measured under the same conditions. For example, if you start with a certain volume of a gas and perform a chemical reaction that doesn't change the temperature or pressure, Avogadro's law lets you find out the volume of another gas formed in the reaction.
Mole-to-Mole Ratio
The mole-to-mole ratio is a valuable concept derived from a balanced chemical equation, indicating the proportions of reactants and products participating in a chemical reaction. This ratio can be used to convert between the amount of moles of one substance to the moles of another substance within the reaction.

To determine the mole-to-mole ratio, you look at the coefficients in a balanced chemical equation. In our example, the ratio between NO鈧� and N鈧侽鈧� is 2:1, meaning two moles of NO鈧� will produce one mole of N鈧侽鈧�. By knowing this ratio and the volume of one of the reactants, in this case, NO鈧�, we can then calculate the expected volume of the product, N鈧侽鈧�.
Balanced Chemical Equation
A balanced chemical equation is essential in stoichiometry because it provides the blueprint for how reactants are transformed into products. In a balanced equation, the number of atoms for each element is the same on both the reactant and the product sides, complying with the law of conservation of mass. Writing and balancing chemical equations is a foundational skill in chemistry.

To balance an equation, you adjust the coefficients in front of the chemical formulas to make sure the same number of atoms for each element appears on both sides of the equation. In our example equation, each side has 2 nitrogen atoms and 4 oxygen atoms, indicating the equation is balanced and ready for further stoichiometric calculations.
Stoichiometry
Stoichiometry is a section of chemistry that involves calculating the quantities of reactants and products involved in a chemical reaction. It is based on a balanced chemical equation and how the elements and compounds are related to each other by the mole-to-mole ratios that the equation provides. Think of it as a recipe for chemicals: knowing the proportions and amounts of ingredients you start with, you can predict the amount of each product you end with.

Using stoichiometry, you can solve various types of problems, including finding the volume of gas produced or required in a reaction, as demonstrated in the original exercise. Once you understand the mole-to-mole ratios and Avogadro's law, you can apply this knowledge to various stoichiometric calculations to predict and understand the outcomes of chemical reactions.

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Most popular questions from this chapter

Which of the following statements is(are) true? For the false statements, correct them. a. At constant temperature, the lighter the gas molecules, the faster the average velocity of the gas molecules. b. At constant temperature, the heavier the gas molecules, the larger the average kinetic energy of the gas molecules. c. A real gas behaves most ideally when the container volume is relatively large and the gas molecules are moving relatively quickly. d. As temperature increases, the effect of interparticle interactions on gas behavior is increased. e. At constant \(V\) and \(T\), as gas molecules are added into a container, the number of collisions per unit area increases resulting in a higher pressure. f. The kinetic molecular theory predicts that pressure is inversely proportional to temperature at constant volume and moles of gas.

A glass vessel contains \(28 \mathrm{~g}\) of nitrogen gas. Assuming ideal behavior, which of the processes listed below would double the pressure exerted on the walls of the vessel? a. Adding \(28 \mathrm{~g}\) of oxygen gas b. Raising the temperature of the container from \(-73^{\circ} \mathrm{C}\) to \(127^{\circ} \mathrm{C}\) c. Adding enough mercury to fill one-half the container d. Adding \(32 \mathrm{~g}\) of oxygen gas e. Raising the temperature of the container from \(30 .{ }^{\circ} \mathrm{C}\) to \(60 .{ }^{\circ} \mathrm{C}\)

Urea \(\left(\mathrm{H}_{2} \mathrm{NCONH}_{2}\right)\) is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of ammonia and carbon dioxide: $$ 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \underset{\text { Pressure }}{\text { Heat }}{\mathrm{H}}_{2} \mathrm{NCONH}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(g) $$ Ammonia gas at \(223^{\circ} \mathrm{C}\) and \(90 .\) atm flows into a reactor at a rate of \(500 . \mathrm{L} / \mathrm{min}\). Carbon dioxide at \(223^{\circ} \mathrm{C}\) and 45 atm flows into the reactor at a rate of \(600 . \mathrm{L} / \mathrm{min} .\) What mass of urea is produced per minute by this reaction assuming \(100 \%\) yield?

Consider two separate gas containers at the following conditions: $$ \begin{array}{|ll|} \text { Container A } & \text { Container B } \\ \text { Contents: } \mathrm{SO}_{2}(g) & \text { Contents: unknown gas } \\ \text { Pressure }=P_{\mathrm{A}} & \text { Pressure }=P_{\mathrm{B}} \\ \text { Moles of gas }=1.0 \mathrm{~mol} & \text { Moles of gas }=2.0 \mathrm{~mol} \\ \text { Volume }=1.0 \mathrm{~L} & \text { Volume }=2.0 \mathrm{~L} \\ \text { Temperature }=7^{\circ} \mathrm{C} & \text { Temperature }=287^{\circ} \mathrm{C} \\ \hline \end{array} $$ How is the pressure in container \(\mathrm{B}\) related to the pressure in container \(\mathrm{A}\) ?

Suppose two 200.0-L tanks are to be filled separately with the gases helium and hydrogen. What mass of each gas is needed to produce a pressure of \(2.70 \mathrm{~atm}\) in its respective tank at \(24^{\circ} \mathrm{C} ?\)
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