Question

In the following reaction, \(2.35 \mathrm{~mol}\) of \(\mathrm{NH}_{3}\) reacts with \(2.75 \mathrm{~mol}\) of \(\mathrm{O}_{2}\); how many moles of water form? $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$

Short answer

3.3 moles of water form

Step-by-step solution

- Write Down the Balanced Chemical Equation

Given the balanced chemical equation: \[ 4 \mathrm{NH}_{3}(g) + 5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g) + 6 \mathrm{H}_{2} \mathrm{O}(l) \]

- Identify the Mole Ratio

From the balanced equation, note the mole ratio between the reactants and the products. This ratio is: \[ 4 \mathrm{NH}_{3} : 5 \mathrm{O}_{2} : 6 \mathrm{H}_{2} \mathrm{O} \]

- Determine the Limiting Reactant

Calculate the moles of water that each reactant can produce to identify the limiting reactant. For \( \mathrm{NH}_{3} \): \[ \frac{2.35 \text{ mol } \mathrm{NH}_{3}}{4 \text{ mol } \mathrm{NH}_{3}} \times 6 \text{ mol } \mathrm{H}_{2} \mathrm{O} = 3.525 \text{ mol } \mathrm{H}_{2} \mathrm{O} \] For \( \mathrm{O}_{2} \): \[ \frac{2.75 \text{ mol } \mathrm{O}_{2}}{5 \text{ mol } \mathrm{O}_{2}} \times 6 \text{ mol } \mathrm{H}_{2} \mathrm{O} = 3.3 \text{ mol } \mathrm{H}_{2} \mathrm{O} \]

- Identify the Limiting Reactant

The limiting reactant is the one that produces the least amount of water. In this case, \( \mathrm{O}_{2} \) produces 3.3 moles of water, which is less than the 3.525 moles produced by \( \mathrm{NH}_{3} \). Therefore, \( \mathrm{O}_{2} \) is the limiting reactant.

- Calculate the Total Moles of Water Formed

Since \( \mathrm{O}_{2} \) is the limiting reactant, the number of moles of water formed is based on \( \mathrm{O}_{2} \). Therefore, the moles of water formed are: \[ 3.3 \text{ mol } \mathrm{H}_{2} \mathrm{O} \]

Key concepts

stoichiometry

Stoichiometry is the study of the quantitative relationships in chemical reactions. In simple terms, it helps us understand and calculate how much of each substance is involved in a reaction. This is achieved using the balanced chemical equation as the basis.

By using stoichiometry, we can determine how much product will form from given amounts of reactants and how much reactant is needed to form a specific amount of product. For example, in our exercise, stoichiometry allows us to calculate the number of moles of water formed when certain amounts of ammonia (NH₃) and oxygen (O₂) react.

The balanced chemical equation provides the necessary ratios between reactants and products to make these calculations. This makes stoichiometry a powerful tool in chemistry for predicting the outcomes of reactions.

balanced chemical equation

A balanced chemical equation represents a chemical reaction with the same number of atoms of each element on both sides of the equation. It ensures that matter is neither created nor destroyed during the reaction. In other words, it follows the Law of Conservation of Mass.

For example, the equation in our exercise is:

\[ 4 \mathrm{NH}_{3}(g) + 5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g) + 6 \mathrm{H}_{2} \mathrm{O}(l) \]

This equation shows that for every 4 moles of ammonia reacting, 5 moles of oxygen are consumed, producing 4 moles of nitric oxide (NO) and 6 moles of water. Balancing equations is crucial for accurate stoichiometric calculations because it provides the correct ratios of reactants and products.

mole ratio

The mole ratio is the ratio of the amounts, in moles, of any two compounds involved in a chemical reaction. It comes directly from the coefficients of the balanced chemical equation.

In our exercise, the balanced equation gives us the mole ratios:

\[ 4 \mathrm{NH}_{3} : 5 \mathrm{O}_{2} : 6 \mathrm{H}_{2} \mathrm{O} \]

This means for every 4 moles of NH₃, we need 5 moles of O₂ to produce 6 moles of H₂O. We can use these ratios to calculate how many moles of a product are formed from a given amount of reactant or how much of one reactant is needed to react with a specific amount of another reactant. Using these ratios simplifies the process of determining the quantities involved in a reaction.

limiting reactant calculation

The limiting reactant is the substance that is entirely consumed first in a chemical reaction, thereby limiting the amount of product that can be formed.

To find the limiting reactant, we compare the amounts of products each reactant can produce.

In our exercise:

• NH₃ can produce: \[ \frac{2.35 \text{ mol NH}_{3}}{4 \text{ mol NH}_{3}} \times 6 \text{ mol H}_{2} \mathrm{O} = 3.525 \text{ mol H}_{2} \mathrm{O} \]
• O₂ can produce: \[ \frac{2.75 \text{ mol O}_{2}}{5 \text{ mol O}_{2}} \times 6 \text{ mol H}_{2} \mathrm{O} = 3.3 \text{ mol H}_{2} \mathrm{O} \]

Since O₂ produces the least amount of water (3.3 moles), it is the limiting reactant. This determines the total amount of water formed in the reaction.

chemical reaction

A chemical reaction is a process where substances, known as reactants, are transformed into different substances called products. During a chemical reaction, bonds between atoms are broken and new bonds are formed, creating new substances with different properties.

Chemical reactions are written using chemical equations that include the reactants and products and often their states (solid, liquid, gas, aqueous).

In the given exercise, the reaction:

\[ 4 \mathrm{NH}_{3}(g) + 5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g) + 6 \mathrm{H}_{2} \mathrm{O}(l) \]

shows how ammonia (NH₃) and oxygen (O₂) react to form nitric oxide (NO) and water (H₂O). The balanced equation allows us to understand the quantitative relationships between the reactants and products, making it a fundamental tool in chemical studies.