Question

Urea, \(\mathrm{NH}_{2} \mathrm{CONH}_{2},\) is a nitrogen fertilizer that is manufactured from ammonia and carbon dioxide. $$ 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ What volume of ammonia at \(25^{\circ} \mathrm{C}\) and \(3.00 \mathrm{~atm}\) is needed to produce \(908 \mathrm{~g}\) ( \(2 \mathrm{lb}\) ) of urea?

Short answer

246.5 L of ammonia is needed.

Step-by-step solution

Determine Moles of Urea

First, find the molar mass of urea, \( \mathrm{NH}_2 \mathrm{CONH}_2 \). The molar mass is the sum of the atomic masses of all atoms in one molecule: \( 2 \times 14.01 + 4 \times 1.01 + 12.01 + 16.00 = 60.06 \text{ g/mol} \). To find the number of moles of urea produced, use the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Plug in the values: \( \frac{908 \text{ g}}{60.06 \text{ g/mol}} \approx 15.12 \text{ moles} \).

Determine Moles of Ammonia Required

According to the balanced chemical equation, 2 moles of \( \mathrm{NH}_3 \) are required to produce 1 mole of urea. Therefore, the moles of \( \mathrm{NH}_3 \) needed are \( 2 \times 15.12 \approx 30.24 \) moles.

Apply the Ideal Gas Law

Use the Ideal Gas Law to find the volume of \( \mathrm{NH}_3 \), where the formula is \( PV = nRT \). Given: \( P = 3.00 \text{ atm}, n = 30.24 \text{ moles}, T = 298 \text{ K} \) (which is \( 25^{\circ}C + 273 = 298 \text{ K} \)), and \( R = 0.0821 \text{ L atm/mol K} \). Rearrange to find \( V = \frac{nRT}{P} \).

Calculate the Volume of Ammonia

Insert the known values into the equation from Step 3: \( V = \frac{30.24 \times 0.0821 \times 298}{3.00} \). Calculate: \( V \approx 246.5 \text{ L} \).

Key concepts

Molar Mass Calculation

Understanding molar mass calculation is crucial for solving chemical problems. The molar mass of a compound is the sum of the atomic masses of all the atoms in its molecular formula. For example, urea, indicated as \( \mathrm{NH}_2 \mathrm{CONH}_2 \), consists of nitrogen, carbon, oxygen, and hydrogen atoms. To calculate its molar mass, we add the molar masses of these elements:

• Nitrogen (\( \mathrm{N} \)): 2 atoms × 14.01 g/mol = 28.02 g/mol
• Hydrogen (\( \mathrm{H} \)): 4 atoms × 1.01 g/mol = 4.04 g/mol
• Carbon (\( \mathrm{C} \)): 1 atom × 12.01 g/mol = 12.01 g/mol
• Oxygen (\( \mathrm{O} \)): 1 atom × 16.00 g/mol = 16.00 g/mol

This yields a total molar mass of 60.06 g/mol. With this, you can convert a given mass of a substance into moles using the formula: \[\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\]This is essential for progressing in chemical calculations.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. The reaction we focus on here is the production of urea from ammonia and carbon dioxide:\[2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(aq)+\mathrm{H}_{2} \mathrm{O}(l)\]In this equation, the reactants \( \mathrm{NH}_3 \) and \( \mathrm{CO}_2 \) undergo a reaction to produce urea \( \mathrm{NH}_2 \mathrm{CONH}_2 \) and water \( \mathrm{H}_2 \mathrm{O} \). Understanding chemical reactions allows us to determine how much reactant is needed or how much product we can obtain from a reaction, a process explored through stoichiometry.

Stoichiometry

Stoichiometry is the heart of chemistry that explains the quantitative relationships between reactants and products in a chemical reaction. By interpreting a balanced chemical equation, like the one for urea production, you can determine the proportion of each reactant needed or the amount of product formed.In this example, the reaction equation tells us that 2 moles of \( \mathrm{NH}_3 \) react with 1 mole of \( \mathrm{CO}_2 \) to produce 1 mole of urea \( \mathrm{NH}_2 \mathrm{CONH}_2 \). Thus, if you need to make 15.12 moles of urea, you would require:

• 2 × 15.12 moles = 30.24 moles of \( \mathrm{NH}_3 \)

Understanding stoichiometry is critical for predicting the amounts of substances consumed and produced in reactions.

Gas Laws

Gas laws describe the behavior of gases and are essential for understanding reactions involving gaseous substances like ammonia \( \mathrm{NH}_3 \). One fundamental gas law is the Ideal Gas Law, expressed as:\[\]\[PV = nRT\]\[\]Here:

• \( P \) is the pressure of the gas
• \( V \) is its volume
• \( n \) represents the number of moles
• \( R \) is the gas constant (0.0821 \( \text{L atm/mol K} \))
• \( T \) is the temperature in Kelvin

By rearranging the equation to solve for \( V \) (volume), you can determine the volume of gas needed. In our ammonia example, if you have 30.24 moles at a temperature of 298 K and a pressure of 3.00 atm, you can calculate the volume needed for the reaction.

Ammonia Production

Ammonia \( \mathrm{NH}_3 \) is a key component in the production of fertilizers like urea. It is often produced through the Haber process, which combines nitrogen and hydrogen gases under high pressure and temperature in the presence of a catalyst.In industrial settings, ammonia is then reacted with carbon dioxide to produce urea. This is a crucial reaction in agriculture since urea provides a rich nitrogen source for plants. Efficient production and utilization of ammonia are both economically and environmentally significant, ensuring that crops have enough nutrients while minimizing waste and emissions.