Question

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 \(\mathrm{Lof} \mathrm{H}_{2} .\) 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

The volume of ammonia gas produced when all of the nitrogen and hydrogen gases are consumed at a constant temperature and pressure is \(2.4\,\mathrm{L\, NH}_{3}\).

Step-by-step solution

Write down the balanced chemical equation

The balanced chemical equation for the reaction between nitrogen and hydrogen gases to form ammonia gas is given by: \[ \mathrm{N}_{2}(g) + 3 \mathrm{H}_{2}(g) \longrightarrow 2\mathrm{NH}_{3}(g) \]

Evaluate if the reaction is under constant temperature and pressure

We are given that the temperature and pressure are constant during the reaction, which means we can use the stoichiometric coefficients directly to calculate the volume of ammonia gas.

Calculate the number of moles of reactants

Under constant temperature and pressure, the ratio of volumes of gases in a reaction is equal to the ratio of the number of moles. In other words, volume ratios are equivalent to mole ratios. Therefore, we can assign mole ratios to the provided volumes of nitrogen and hydrogen gases: \[ 1.2\,\mathrm{L\, N}_{2}\times\frac{1\,\text{mol N}_2}{1\,\mathrm{L\, N}_{2}} = 1.2\,\text{mol N}_{2}\] \[3.6\,\mathrm{L\,H}_{2}\times\frac{1\,\text{mol H}_{2}}{3.6\,\mathrm{L\, H}_{2}} = 1\,\text{mol H}_{2}\]

Find the limiting reactant

We can find out if there is any limiting reactant by comparing the mole ratios of reactants in the balanced equation (\(1\:\mathrm{N}_{2}\) to \(3\:\mathrm{H}_{2}\)) to the given moles of reactants. The required ratio of moles of \(\mathrm{H}_{2}\) to \(\mathrm{N}_{2}\) is \(3:1\), but we have a ratio of \(1:1.2\), which indicates that there is not enough \(\mathrm{H}_{2}\) for all of the \(\mathrm{N}_2\) to react. Therefore, \(\mathrm{H}_{2}\) is the limiting reactant.

Calculate the volume of ammonia gas produced

Now that we have identified the limiting reactant (\(\mathrm{H}_{2}\)), we can use the stoichiometry from the balanced chemical equation to determine the volume of ammonia produced. In the equation, 3 moles of \(\mathrm{H}_{2}\) react to produce 2 moles of \(\mathrm{NH}_{3}\), so we can calculate the moles of ammonia produced: \[1\,\text{mol}\,\mathrm{H}_{2}\times\frac{2\,\text{mol}\,\mathrm{NH}_{3}}{3\,\text{mol}\,\mathrm{H}_{2}} = \frac{2}{3}\,\text{mol}\,\mathrm{NH}_{3}\] Since the moles and volumes are directly proportional in this case, we can now calculate the volume of ammonia produced: \[\frac{2}{3}\,\mathrm{mol\,NH}_{3}\times\frac{3.6\,\mathrm{L\,H}_{2}}{1\,\mathrm{mol\,H}_{2}} = 2.4\,\mathrm{L\, NH}_{3}\]

State the final answer

The volume of ammonia gas produced when all of the nitrogen and hydrogen gases are consumed at a constant temperature and pressure is 2.4 L.

Key concepts

Chemical Reaction Balancing

Understanding the fundamental aspects of balancing chemical reactions is essential for any stoichiometry calculation. A balanced chemical equation ensures that the same number of atoms of each element are present on both sides of the equation, adhering to the conservation of mass.

For instance, the reaction of nitrogen and hydrogen to form ammonia is represented by the balanced equation: \[N_{2}(g) + 3H_{2}(g) \longrightarrow 2NH_{3}(g)\]This equation tells us that one molecule of nitrogen gas reacts with three molecules of hydrogen gas to produce two molecules of ammonia gas. Balancing chemical equations is the first step in any stoichiometry problem because it allows us to work out the quantitative relationships between reactants and products.

Limiting Reactant Calculation

In chemical reactions, the limiting reactant is the substance that is completely consumed first, thereby determining the maximum amount of product that can be formed. To identify it, one must compare the mole ratios of the available reactants to the ratios needed by the balanced equation.

For our nitrogen and hydrogen example, even though nitrogen seems to be in lesser volume, a stoichiometric analysis shows that hydrogen gas is actually the limiting reactant. This implies that the production of ammonia will stop when all of the hydrogen is used up. Understanding which reactant limits the reaction is paramount in predicting the amounts of products formed.

Volume-to-Mole Ratio

The relation between the volume and amount (in moles) of a gas at constant temperature and pressure is vital to stoichiometry in gas-phase reactions. Avogadro's law indicates that equal volumes of gases, at the same temperature and pressure, have the same number of molecules. Therefore, we can use volume in lieu of moles as a direct representative of quantity when these conditions are applied.

In our exercise, we converted the volumes of nitrogen and hydrogen to moles using unity because the volume-to-mole ratio is 1:1 under constant conditions. This simplicity is a powerful tool in solving gas-phase stoichiometry problems without explicitly converting volumes to moles and vice versa when the conditions remain unchanged.

Avogadro's Law

Avogadro's law is a cornerstone of gas laws and stoichiometry. It states that the volume of a gas is directly proportional to the number of moles, provided the temperature and pressure are constant. Essentially, this means that one mole of any gas occupies the same volume as one mole of any other gas under the same conditions.

In practical terms, this law allows us to relate the volumes of reacting gases to the number of moles, which leads to a simple ratio-based approach to stoichiometric calculations. Applying Avogadro's law in our exercise permits us to predict that 2.4 L of ammonia gas will be produced when 1.2 L of nitrogen gas and 3.6 L of hydrogen gas react completely at constant temperature and pressure.