Problem 101 - Chemistry The Central Science

Question

A mixture of \(\mathrm{N}_{2}(g)\) and \(\mathrm{H}_{2}(g)\) reacts in a closed container to form ammonia, \(\mathrm{NH}_{3}(g)\). The reaction ceases before either reactant has been totally consumed. At this stage \(3.0 \mathrm{~mol} \mathrm{~N}_{2}, 3.0\) \(\mathrm{mol} \mathrm{H}_{2},\) and \(3.0 \mathrm{~mol} \mathrm{NH}_{3}\) are present. How many moles of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) were present originally?

Step-by-Step Solution

Answer

The original moles of N2 and H2 present in the container were 2.0 moles of N2 and 3.0 moles of H2.

Step 1: Write down the balanced equation

The balanced equation for the reaction is: \[N_2(g) + 3H_2(g) \rightarrow 2NH_3(g).\]

Step 2: Write down the given data

We know that after the reaction has ceased, there is: - 3.0 moles of N2 - 3.0 moles of H2 - 3.0 moles of NH3

Step 3: Calculate the amount of N2 and H2 reacted

Let's assume that x moles of N2 have reacted and y moles of H2 have reacted. From the balanced equation, we know that: - 1 mole of N2 reacts with 3 moles of H2 to form 2 moles of NH3 - x moles of N2 will react with 3x moles of H2 to form 2x moles of NH3 - Now we can sum the moles of each reactant and product, according to the given data: - N2: 3.0 (final) = x (reacted) + original (unknown) - H2: 3.0 (final) = 3x (reacted) + original (unknown) - NH3: 3.0 (final) = 2x (reacted)

Step 4: Calculate the amount of NH3 formed from N2 and H2

From the third equation, we find the amount of NH3 formed by the reaction as 3/2 times the moles of N2 reacted: - 3.0 (final) = 2x (reacted) - x = 1.5 moles

Step 5: Calculate the amount of N2 and H2 reacted

Now let's plug in the value of x into the equations for N2 and H2: - N2: 3.0 (final) = 1.5 (reacted) + original N2 - H2: 3.0 (final) = 3(1.5) (reacted) + original H2

Step 6: Calculate the original moles of N2 and H2

Solving for the original moles of N2 and H2: - Original N2 = 3.0 (final) - 1.5 (reacted) = 1.5 moles - Original H2 = 3.0 (final) - 4.5 (reacted) = -1.5 moles However, a negative amount of moles is impossible. Recall that the reaction ceased before either reactant was consumed completely. This indicates that the amount of H2 calculated is incorrect due to the reaction not being carried out to completion. So, we reevaluate the stoichiometry based on the limiting reactant, H2: If the entire 3 moles of H2 were present initially and completely reacted, then 1 mole of H2 will react with 1/3 mole of N2, and therefore 3 moles of H2 will react with 1 mole of N2. This would result in: - Original N2: 3.0 (final) - 1.0 (reacted) = 2.0 moles - Original H2: 3.0 (final) - 0 (reacted) = 3.0 moles Thus, the correct answer is that there were originally 2.0 moles of N2 and 3.0 moles of H2 present.

Key Concepts

Chemical Reactions

When elements or compounds come together, they might undergo a change called a chemical reaction. During this process, the atoms in the reactants rearrange themselves to form new substances known as products. Chemical reactions happen all around us and are essential to life as well as technology.

In a chemical reaction, there are reactants that start the reaction, and products that are formed as a result. Understanding these changes is crucial in stoichiometry, where we analyze the quantities of substances that take part in reactions. In the exercise given, nitrogen \(N_2\) and hydrogen \(H_2\) are reactants combining to form ammonia \(NH_3\).

A crucial aspect of any chemical reaction is the conservation of mass; atoms are neither created nor destroyed during the reaction. They simply rearrange. This principle forms the basis of writing balanced chemical equations, which depict these reactions accurately.

Limiting Reactants

In many chemical reactions, one reactant is often used up before the other. This reactant is called the limiting reactant because it determines how much product can be formed in the reaction. Once the limiting reactant is completely consumed, the reaction stops, even if other reactants are still available.

Understanding limiting reactants is essential for predicting the quantities of products formed in chemical reactions. In the given exercise, we consider the amounts of nitrogen \(N_2\) and hydrogen \(H_2\) available. The reaction stops when either no more reactants can form products or when stoichiometry dictates it.

In our stoichiometric calculations, we assess both nitrogen and hydrogen to identify which one limits the formation of ammonia. In this case, hydrogen acts as the limiting reactant, as we discovered there were challenges to fully react all of \(H_2\) molecules available while \(N_2\) still remained.

Balanced Equations

A balanced equation is a mathematical representation of a chemical reaction where the number of atoms of each element is the same on both sides of the equation. This balance is crucial because it reflects the conservation of mass and ensures the reaction is correctly represented.

In our original exercise, the balanced equation is written as: \[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \] This tells us that one molecule of nitrogen reacts with three molecules of hydrogen to produce two molecules of ammonia.

Balancing equations involves ensuring the same number and type of atoms on both sides. It requires proper coefficient selection, which can often change based on which reactant is in excess or limiting. For example, if there were a different starting amount of reactants, balance would require different coefficients to accurately represent the reaction.

Without balanced equations, it would be challenging to perform accurate stoichiometric calculations, as these equations inform how many molecules or moles of a substance are involved in the reaction.