Question

Billions of pounds of urea, \(\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}\), are produced annually for use as a fertilizer. The principal reaction employed is $$2 \mathrm{NH}_{3}+\mathrm{CO}_{2} \rightarrow \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}+\mathrm{H}_{2} \mathrm{O}$$ By assuming unlimited amounts of \(\mathrm{CO}_{2}\), how many moles of urea can be produced from each of the following amounts of \(\mathrm{NH}_{3}\) ? a. \(2 \mathrm{~mol} \mathrm{NH}_{3}\) b. \(0.45 \mathrm{~mol} \mathrm{NH}_{3}\) c. \(10 \mathrm{~g} \mathrm{NH}_{3}\) d. \(2.0 \mathrm{~kg} \mathrm{NH}_{3}\)

Short answer

a. 1 mole of urea, b. 0.225 moles of urea, c. 0.294 moles of urea, d. 58.72 moles of urea

Step-by-step solution

- Understand the Chemical Reaction

Analyze the chemical equation given: \(2 \mathrm{NH}_{3}+\mathrm{CO}_{2} \rightarrow \mathrm{CO}(\mathrm{NH}_{2})_{2}+\mathrm{H}_{2}O\). The equation implies that 2 moles of ammonia (\(\mathrm{NH}_{3}\)) react with 1 mole of carbon dioxide (\(\mathrm{CO}_{2}\)) to produce 1 mole of urea (\(\mathrm{CO}(\mathrm{NH}_{2})_{2}\)) and 1 mole of water (\(\mathrm{H}_{2}O\)).

- Mole-Mole Relationship

Use the mole-mole relationship from the chemical reaction to determine how many moles of urea can be produced from ammonia. As the equation states, 2 moles of \(\mathrm{NH}_{3}\) yield 1 mole of urea.

- Calculate Moles of Urea from 2 moles of NH3

Given that there are 2 moles of \(\mathrm{NH}_{3}\), and understanding the stoichiometry of the reaction, straightforwardly determine that 2 moles of ammonia will produce 1 mole of urea (1:1 ratio, according to the reaction).

- Calculate Moles of Urea from 0.45 moles of NH3

From 0.45 moles of \(\mathrm{NH}_{3}\), use the ratio from the balanced equation. Divide the number of moles of \(\mathrm{NH}_{3}\) by 2 to find how many moles of urea will be produced: \(0.45 \, \text{moles of} \, \mathrm{NH}_{3} \div 2 = 0.225 \, \text{moles of urea}\).

- Converting Grams to Moles for NH3 (for part c)

To convert grams of ammonia to moles, use the molar mass of ammonia: \(\mathrm{NH}_{3} = 14.01 \, \text{g/mol} (N) + 3 \times 1.01 \, \text{g/mol} (H) = 17.03 \, \text{g/mol}\). Calculate moles of 10g of \(\mathrm{NH}_{3}\): \(10 \, \text{g of} \, \mathrm{NH}_{3} \div 17.03 \, \text{g/mol} = 0.587 \, \text{moles of} \, \mathrm{NH}_{3}\).

- Calculate Moles of Urea from 10g of NH3

Use the moles obtained from Step 5 and the mole-mole relationship from the balanced equation to find the moles of urea: \(0.587 \, \text{moles of} \, \mathrm{NH}_{3} \div 2 = 0.294 \, \text{moles of urea}\).

- Converting Kilograms to Moles for NH3 (for part d)

Convert kg to g: \(2.0 \, \text{kg} = 2000 \, \text{g}\). Use the molar mass of ammonia to convert to moles: \(2000 \, \text{g} \div 17.03 \, \text{g/mol} = 117.44 \, \text{moles of} \, \mathrm{NH}_{3}\).

- Calculate Moles of Urea from 2.0 kg of NH3

Apply the mole-mole relationship to find the moles of urea from the moles of ammonia found in Step 7: \(117.44 \, \text{moles of} \, \mathrm{NH}_{3} \div 2 = 58.72 \, \text{moles of urea}\).

Key concepts

Chemical Equation Balancing

Understanding how to balance chemical equations is pivotal for studying stoichiometry in chemical reactions. It ensures the law of conservation of mass is followed, meaning that the number of atoms for each element is the same on both sides of the equation.

For instance, the given reaction for producing urea is balanced as follows: for every 2 moles of ammonia (\textbf{NH}\(_3\)) reacted, 1 mole of carbon dioxide (\textbf{CO}\(_2\)) is used, resulting in 1 mole of urea and 1 mole of water (\textbf{H}\(_2\)\textbf{O}). Here, balancing helps to visually represent the mole-to-mole ratio among the reactants and the products, which is essential for predicting outcomes of a chemical reaction.

Importance in Calculations

Without a balanced equation, it would be impossible to calculate the quantities of reactants needed or products formed, as seen in the provided exercise where ammonia's quantity determines the amount of urea produced. Balancing chemical equations is not just a formality but a fundamental method to approach stoichiometric problems effectively.

Mole-Mole Relationship

The heart of stoichiometry lies within the mole-mole relationships established by a balanced chemical equation. This relationship tells us how many moles of one substance will react with or produce a certain number of moles of another substance.

In our exercise, we look at the mole-mole ratio between ammonia and urea. For every 2 moles of ammonia consumed, 1 mole of urea is produced. This means if we start with fewer than 2 moles of ammonia, we'll produce less than 1 mole of urea, highlighting how this ratio serves as a conversion factor between reactants and products.

Practical Application

When solving problems, like for instance b. 0.45 moles of \textbf{NH}\(_3\), you divide this number by 2, as dictated by the ratio, to determine 0.225 moles of urea will be formed. This step illustrates the direct application of the mole-mole relationship in stoichiometric calculations, enabling precise predictions in quantitative analysis.

Molar Mass Calculation

Calculating molar mass is a fundamental step when dealing with conversions between mass and number of moles in stoichiometry. The molar mass is defined as the mass of one mole of a substance and is usually expressed in grams per mole (g/mol).

For chemical reactions, knowing the molar mass allows for the conversion of a given mass of a substance into moles, which then can be used within the stoichiometric framework of the reaction. In Step 5 of the solution, we calculate the molar mass of ammonia (\textbf{NH}\(_3\)) by adding the atomic masses of one nitrogen atom (14.01 g/mol) and three hydrogen atoms (3 times 1.01 g/mol), resulting in 17.03 g/mol for ammonia.

Interpreting in Calculations

Once the molar mass is known, it can be used to convert the mass of a substance to moles. For part c of the exercise, 10 grams of ammonia is divided by its molar mass to find the moles of ammonia present, which is essential for determining the amount of urea that can be produced. Efficiently understanding and applying molar mass calculations is indispensable for mastering stoichiometry.