Question

About \(8.0 \times 10^{6}\) tons of urea \(\left[\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\right]\) is used annually as a fertilizer. The urea is prepared at \(200^{\circ} \mathrm{C}\) and under high-pressure conditions from carbon dioxide and ammonia (the products are urea and steam). Calculate the volume of ammonia (in liters) measured at 150 atm needed to prepare 1.0 ton of urea.

Short answer

The volume of ammonia needed is approximately 8613.33 liters.

Step-by-step solution

Understanding the Problem

We need to find the volume of ammonia in liters at 150 atm required to produce 1.0 ton of urea.

Write the Balanced Chemical Equation

The chemical reaction for producing urea from carbon dioxide and ammonia is as follows:\[ 2 ext{ NH}_3 + ext{ CO}_2 ightarrow ext{ NH}_2 ext{CONH}_2 + ext{ H}_2 ext{O} \]

Calculate the Molar Mass of Urea

The molecular formula of urea is \(\mathrm{(NH}_2)_2\mathrm{CO}\). Calculate its molar mass:- N: 14.01 g/mol (since there are two nitrogen atoms, multiply by 2)- H: 1.01 g/mol (since there are four hydrogen atoms, multiply by 4)- C: 12.01 g/mol- O: 16.00 g/molSum these to get the molar mass of urea:\[2 \times 14.01 + 4 \times 1.01 + 12.01 + 16.00 = 60.06 \text{ g/mol} \]

Convert Tons to Grams

1 ton is equal to 1,000,000 grams. Therefore, 1.0 ton of urea is 1,000,000 grams.

Calculate Moles of Urea

Using the molar mass of urea, calculate the number of moles in 1.0 ton of urea:\[\text{Moles of urea} = \frac{1,000,000 \text{ g}}{60.06 \text{ g/mol}} \approx 16,649.45 \text{ moles}\]

Determine Moles of Ammonia Required

According to the balanced equation, 2 moles of NH₃ are needed for every mole of urea. Therefore:\[\text{Moles of NH}_3 = 2 \times 16,649.45 \approx 33,298.9 \text{ moles}\]

Apply Ideal Gas Law to Find Volume of Ammonia

Using the ideal gas law, \( PV = nRT \), solve for \(V\) to find the volume. Assume \(R = 0.0821 \text{ L atm/mol K}\) and the temperature is \(200^{\circ} \text{C} = 473 \text{K}\).Rearrange the equation:\[ V = \frac{nRT}{P} \]Substitute the values:\[ V = \frac{33,298.9 \times 0.0821 \times 473}{150} \approx 8613.33 \text{ liters}\]

Conclusion

The volume of ammonia required to produce 1.0 ton of urea at 150 atm is approximately 8613.33 liters.

Key concepts

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show the reactants turning into products through a series of symbols and formulas. For the synthesis of urea, the chemical equation is crucial because it identifies the necessary starting materials and the quantities required for the reaction. In this particular case, the balanced equation is:
\[2 \text{ NH}_3 + \text{ CO}_2 \rightarrow \text{ NH}_2\text{CONH}_2 + \text{ H}_2\text{O} \]
Here, two molecules of ammonia \(NH_3\) and one molecule of carbon dioxide \(CO_2\) react to form one molecule of urea \(NH_2CONH_2\) and one molecule of water \(H_2O\). Balancing is important because it respects the Law of Conservation of Mass, ensuring the same number of atoms of each element on both sides of the equation.

Molecular Mass Calculation

Calculating the molecular mass involves summing the masses of all the atoms in a molecule. This is essential in conversions between mass and moles, a common task in stoichiometry. For urea, its formula is \((NH_2)_2CO\), and its molar mass is derived from the atomic masses of nitrogen (N), hydrogen (H), carbon (C), and oxygen (O).
- N: 14.01 g/mol (2 atoms, so \(2 \times 14.01\))- H: 1.01 g/mol (4 atoms, so \(4 \times 1.01\))- C: 12.01 g/mol- O: 16.00 g/mol
Adding these, the molar mass of urea is:\[2 \times 14.01 + 4 \times 1.01 + 12.01 + 16.00 = 60.06 \text{ g/mol} \]
This value is critical for converting grams to moles when working with chemical reactions.

Ideal Gas Law

The ideal gas law is a pivotal principle in chemistry for relating the pressure, volume, temperature, and moles of a gas. It is expressed by the equation:
\[ PV = nRT \]
Where:

• \(P\) is the pressure of the gas (in atm)
• \(V\) is the volume of the gas (in liters)
• \(n\) is the number of moles of the gas
• \(R\) is the ideal gas constant (0.0821 L atm/mol K)
• \(T\) is the temperature (in Kelvin)

Applying this law allows for the calculation of any one of these variables if the others are known. In this exercise, it helps determine the volume of ammonia needed, given specific pressure and temperature conditions.

Stoichiometry

Stoichiometry deals with the calculations of reactants and products in chemical reactions, allowing chemists to predict the amounts of substances consumed and produced. It is grounded in the mole concept and balanced chemical equations.

In the synthesis of urea, stoichiometry tells us that two moles of ammonia are required to produce one mole of urea, making it possible to calculate the number of moles of ammonia required for any given amount of urea.
For example, using the balanced equation from earlier, if 1.0 ton (1,000,000 grams) of urea corresponds to approximately 16,649.45 moles of urea, then twice as many moles of ammonia are needed, which can be calculated as:\[2 \times 16,649.45 = 33,298.9 \text{ moles} \]
This understanding is essential when scaling up reactions for industrial production.