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Chapter 10: Problem 28

Nitrogen and hydrogen gases react to form ammonia gas as follows: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ At a certain temperature and pressure, \(1.2 \mathrm{~L}\) of \(\mathrm{N}_{2}\) reacts with \(3.6\) L of \(H_{2}\). If all the \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) are consumed, what volume of \(\mathrm{NH}_{3}\), at the same temperature and pressure, will be produced?

Short Answer

Expert verified
The volume of ammonia gas (NH3) produced when 1.2 L of N2 reacts with 3.6 L of H2 will be 2.4 L at the same temperature and pressure.

Step by step solution

01

Identifying the limiting reactant

We can start by identifying the limiting reactant in this reaction. To do this, we can compare the volume ratios of N2 and H2 used in the reaction. The balanced chemical equation is: \[ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \] The ratio of N2 to H2 should be 1:3, according to the stoichiometry. Now let's check the ratio given in the problem: \[ \frac{1.2 \,\textnormal{L} \,\, \mathrm{N}_{2}}{3.6 \,\textnormal{L} \,\, \mathrm{H}_{2}}=\frac{1}{3} \] Since the ratio matches the stoichiometry of the reaction, both reactants are consumed completely, and there is no limiting reactant or excess reactant.
02

Determining the volume of ammonia gas produced

Now that we know that all the N2 and H2 are consumed, we can determine the volume of ammonia gas (NH3) produced. According to the stoichiometry of the reaction: \[ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) \] For each mole of N2 that reacts, 2 moles of NH3 are produced. We can use this ratio to find the volume of NH3 produced. Since the volume ratio of gases at the same temperature and pressure is equal to the mole ratio, we can write: \[ \frac{V_{\mathrm{NH}_{3}}}{V_{\mathrm{N}_{2}}}=\frac{2}{1} \] Now, we can find the volume of NH3 produced: \[ V_{\mathrm{NH}_{3}} = 2 \times V_{\mathrm{N}_{2}} \] Given that the volume of N2 is 1.2 L, we can calculate the volume of NH3: \[ V_{\mathrm{NH}_{3}} = 2 \times 1.2 \,\textnormal{L} \]
03

Calculating the volume of ammonia gas produced

Now we can perform the calculation: \[ V_{\mathrm{NH}_{3}} = 2 \times 1.2 \,\textnormal{L} = 2.4 \,\textnormal{L} \] Thus, if all the N2 and H2 are consumed, 2.4 L of ammonia gas (NH3) will be produced at the same temperature and pressure.

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Key Concepts

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

Limiting Reactant
In any chemical reaction, the limiting reactant is the substance that is completely consumed first, determining the maximum amount of product that can be formed. In reactions involving gases, volume can be used directly to determine the limiting reactant as long as all gases are measured under the same conditions of temperature and pressure.
In our exercise, we have nitrogen gas (\( \text{N}_2 \)) reacting with hydrogen gas (\( \text{H}_2 \)) to produce ammonia (\( \text{NH}_3 \)). According to the balanced equation:
  • 1 \( \text{N}_2 \) molecule reacts with 3 \( \text{H}_2 \) molecules to produce 2 \( \text{NH}_3 \) molecules.

From the problem, we see that the ratio of \( \text{N}_2 \) to \( \text{H}_2 \) provided (1.2 L : 3.6 L) matches the stoichiometric ratio of 1:3 perfectly, meaning neither \( \text{N}_2 \) nor \( \text{H}_2 \) is in excess. Both are limiting each other in a perfectly balanced way. Consequently, all reactant molecules are consumed entirely, without any left over.
Gas Laws
Gas laws are crucial in understanding the behavior of gases during chemical reactions. These laws, such as Boyle's, Charles', and Avogadro's, describe the relationships between the volume, pressure, temperature, and number of moles of a gas.
In reactions involving gases like this one, the ideal gas law becomes very useful:\[ PV = nRT \] This equation relates pressure (P), volume (V), number of moles (n), and temperature (T) with the gas constant (R).
While the detailed calculation using the ideal gas law isn't required here, we rely on the principle that at constant temperature and pressure, the volume occupied by a gas is directly proportional to the number of moles. This is Avogadro's Law, simplifying our problem by allowing us to use the volume ratio directly as a mole ratio.
  • This means if the conditions remain constant, doubling the amount of \( \text{N}_2 \) produces double the volume of \( \text{NH}_3 \) gas.
Chemical Reactions
Chemical reactions involve the reorganization of atoms to transform reactants into products. They can be represented by balanced chemical equations, showing reactant and product ratios.
In this exercise, we explore the reaction of nitrogen (\( \text{N}_2 \)) with hydrogen (\( \text{H}_2 \)) to form ammonia (\( \text{NH}_3 \)). The balanced equation:\[ \text{N}_2(g) + 3 \text{H}_2(g) \rightarrow 2 \text{NH}_3(g) \]describes the stoichiometry of this reaction, indicating that:
  • 1 molecule of \( \text{N}_2 \) reacts with 3 molecules of \( \text{H}_2 \).
  • This process produces 2 molecules of \( \text{NH}_3 \).

Understanding this stoichiometry helps us predict the amounts of products formed when certain amounts of reactants are used. In this problem, for every 1 volume unit of \( \text{N}_2 \), 2 volume units of \( \text{NH}_3 \) are produced. Thus, with 1.2 L of \( \text{N}_2 \), we expect 2.4 L of \( \text{NH}_3 \) as the result.

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Most popular questions from this chapter

Newton had an incorrect theory of gases in which he assumed that all gas molecules repel one another and the walls of their container. Thus, the molecules of a gas are statically and uniformly distributed, trying to get as far apart as possible from one another and the vessel walls. This repulsion gives rise to pressure. Explain why Charles's law argues for the kinetic- molecular theory and against Newton's model.

Which of the following statements best explains why nitrogen gas at STP is less dense than Xe gas at STP? (a) Because Xe is a noble gas, there is less tendency for the Xe atoms to repel one another, so they pack more densely in the gas state. (b) Xe atoms have a higher mass than \(\mathrm{N}_{2}\) molecules. Because both gases at STP have the same number of molecules per unit volume, the Xe gas must be denser. (c) The Xe atoms are larger than \(\mathrm{N}_{2}\) molecules and thus take up a larger fraction of the space occupied by the gas. (d) Because the Xe atoms are much more massive than the \(\mathrm{N}_{2}\) molecules, they move more slowly and thus exert less upward force on the gas container and make the gas appear denser.

A \(4.00-\mathrm{g}\) sample of a mixture of \(\mathrm{CaO}\) and \(\mathrm{BaO}\) is placed in a 1.00-L vessel containing \(\mathrm{CO}_{2}\) gas at a pressure of 730 torr and a temperature of \(25^{\circ} \mathrm{C}\). The \(\mathrm{CO}_{2}\) reacts with the \(\mathrm{CaO}\) and \(\mathrm{BaO}\), forming \(\mathrm{CaCO}_{3}\) and \(\mathrm{BaCO}_{3}\). When the reaction is complete, the pressure of the remaining \(\mathrm{CO}_{2}\) is 150 torr. (a) Calculate the number of moles of \(\mathrm{CO}_{2}\) that have reacted. (b) Calculate the mass percentage of \(\mathrm{CaO}\) in the mixture.

The density of a gas of unknown molar mass was measured as a function of pressure at \(0{ }^{\circ} \mathrm{C}\), as in the table below. (a) Determine a precise molar mass for the gas. Hint: Graph \(d / P\) versus \(P\). (b) Why is \(d / P\) not a constant as a function of pressure? $$ \begin{array}{llllll} \hline \text { Pressure } & & & & & \\ \begin{array}{l} \text { (atm) } \end{array} & 1.00 & 0.666 & 0.500 & 0.333 & 0.250 \\ \text { Density } & & & & & \\ \begin{array}{l} \text { (g/L) } \end{array} & 2.3074 & 1.5263 & 1.1401 & 0.7571 & 0.5660 \\ \hline \end{array} $$

For nearly all real gases, the quantity \(P V / R T\) decreases below the value of 1, which characterizes an ideal gas, as pressure on the gas increases. At much higher pressures, however, \(P V / R T\) increases and rises above the value of 1 . (a) Explain the initial drop in value of \(P V / R T\) below 1 and the fact that it rises above 1 for still higher pressures. (b) The effects we have just noted are smaller for gases at higher temperature. Why is this so?
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