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 \(3.6 \mathrm{~L}\) of \(\mathrm{H}_{2}\). 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 NH3 produced when all the N2 and H2 are consumed, at the same temperature and pressure, is \(2.4 \ \mathrm{L}\).

Step-by-step solution

Identify the Balanced Chemical Equation

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

Determine Mole Ratios

Using the balanced chemical equation, we can determine the mole ratios of the reactants and products. In this case, the mole ratio of N2 to H2 is 1:3, and the mole ratio of N2 to NH3 is 1:2.

Calculate Volume of NH3 Produced

We are given the volumes of N2 and H2 reacting as 1.2 L and 3.6 L, respectively. Based on the mole ratios and using the balanced chemical equation, we can calculate the volume of NH3 produced as follows: First, we need to check which reactant is the limiting reactant. Since the mole ratio of N2 to H2 is 1:3, we can divide the volume of H2 by 3: \[ \frac{3.6 \ \mathrm{L}}{3} = 1.2 \ \mathrm{L} \] Since the volume of N2 is equal to this value, this means that both reactants are completely consumed in the reaction, and we can proceed with the calculation. Using the mole ratio of N2 to NH3 (1:2), we can now calculate the volume of NH3 produced: \[ 1.2 \ \mathrm{L} \times \frac{2 \ \mathrm{moles \ of \ NH}_{3}}{1 \ \mathrm{mole \ of \ N}_{2}} \] This simplifies to: \[ 1.2 \ \mathrm{L} \times 2 = 2.4 \ \mathrm{L} \]

State the Volume of NH3 Produced

Since all the N2 and H2 are consumed, the volume of NH3 produced, at the same temperature and pressure as the reactants, is 2.4 L.

Key concepts

Stoichiometry

Stoichiometry is the art of measuring the elements and is fundamental in understanding chemical reactions. It involves using balanced chemical equations to predict the quantity of reactants and products in a chemical reaction.
One might think of stoichiometry as a recipe in cooking that lays out the precise amount of each ingredient needed to achieve the desired result.
In chemistry, it's about predicting how substances will interact. The core of stoichiometry lies in the mole concept.- Mole Ratios: These are derived from the coefficients shown in a balanced chemical equation. They establish the proportion relationships between reactants and products.
- Balanced Equations: Ensuring the same number of atoms of each element on both sides of a chemical reaction is vital. For our exercise, the equation is balanced as: \[ \mathrm{N}_{2}(g) + 3\ \mathrm{H}_{2}(g) \rightarrow 2\ \mathrm{NH}_{3}(g) \]In the exercise, the mole ratios help us understand that one mole of nitrogen reacts with three moles of hydrogen to produce two moles of ammonia.
This understanding aids in calculating the required amounts to be reacted and produced, just like ensuring you follow a cookie recipe to make exactly a dozen cookies.

Limiting Reactant

In every chemical reaction, one reactant will be consumed first, putting a stop to the reaction. This is known as the limiting reactant. It's like running out of eggs when you're baking; even if you have plenty of flour, you can't make more cakes without eggs.
To find the limiting reactant, compare the mole ratios of the reactants using the volumes or masses provided.- Matching Ratios: When the reactants are given in volumes, as in our exercise, the actual volume ratios should match the mole ratios from the balanced equation.
- Calculation: In the situation where nitrogen and hydrogen gases react, we have 1.2 L of \( \mathrm{N}_{2} \) and 3.6 L of \( \mathrm{H}_{2} \). The ratio is exactly 1:3, matching the expected reduction in the amounts that participate in the reaction.When the reactants are in the exact ratio, as here, both are completely used up, allowing further calculation to determine the extent of products formed, but neither is in excess.

Gas Volume Calculations

Gas volume calculations at constant temperature and pressure can simplify to mole calculations. This is thanks to Avogadro's Law, which states that equal volumes of gases at the same temperature and pressure will contain an equal number of molecules.
In this particular exercise, the conditions allow us to directly use the volumes provided as the mole ratios.- Using Mole Ratio: Once you've identified that the reactants are at the exact proportions required, the calculation of the volume of \( \mathrm{NH}_{3} \) produced can be straightforward.
- Direct Substitution: The volume of \( \mathrm{NH}_{3} \) produced is calculated as 2.4 L because of the direct relationship established through the mole ratio of 1:2 between \( \mathrm{N}_{2} \) and \( \mathrm{NH}_{3} \).The principle of using volume to equate moles is powerful, especially in situations where calculating moles directly isn't feasible.
This allows chemists to predict outcomes of reactions quickly and with accuracy when dealing with gases.