CO2(g) + 2 NH3(g) CO(NH2)2(s) + H2O(g)
What volumes of CO2 and NH3 at 193 atm and 450.°C are needed to produce 2.54 kg of urea, assuming the reaction goes to completion?
Use this procedure in the worked example to solve for moles CO2 and moles NH3 needed.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Then use PV = nRT and the conditions listed to convert moles to volume.
Don't forget to convert T to kelvin.
To answer this question, we need to use the ideal gas law to calculate the volumes of carbon dioxide (CO2) and ammonia (NH3) gas required.
1. Let's first convert the given pressure from atmospheres (atm) to Pascals (Pa). Recall that 1 atm = 101325 Pa.
Given pressure: 193 atm
Converted pressure: 193 atm * 101325 Pa/atm = 1.9549 × 10^7 Pa
2. Next, let's convert the given temperature from Celsius to Kelvin (K). Recall that Kelvin = Celsius + 273.15.
Given temperature: 450.°C
Converted temperature: 450.°C + 273.15 = 723.15 K
3. Now, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, we need to calculate the number of moles of urea (CO(NH2)2) using its molar mass. The molar mass of urea can be found by adding the molar masses of its constituent elements:
Carbon (C): 12.01 g/mol
Oxygen (O): 16.00 g/mol
Nitrogen (N): 14.01 g/mol
Hydrogen (H): 1.01 g/mol
Urea (CO(NH2)2): (12.01 + (2 * 16.00) + (2 * 14.01) + (4 * 1.01)) g/mol = 60.06 g/mol
Given mass of urea: 2.54 kg = 2540 g
Number of moles of urea: 2540 g / 60.06 g/mol = 42.3 mol
4. The balanced equation tells us that 1 mole of carbon dioxide (CO2) reacts with 2 moles of ammonia (NH3) to produce 1 mole of urea. Therefore, the number of moles of CO2 and NH3 needed is the same as the number of moles of urea.
Number of moles of CO2 and NH3: 42.3 mol
5. Now, we can rearrange the ideal gas law equation to solve for volume (V):
V = nRT / P
For CO2:
Volume of CO2 = (42.3 mol * 8.314 J/(mol·K) * 723.15 K) / (1.9549 × 10^7 Pa) (using R = 8.314 J/(mol·K))
For NH3:
Volume of NH3 = (42.3 mol * 8.314 J/(mol·K) * 723.15 K) / (1.9549 × 10^7 Pa) (using R = 8.314 J/(mol·K))
6. Finally, calculate the volumes of CO2 and NH3:
Volume of CO2 = (42.3 mol * 8.314 J/(mol·K) * 723.15 K) / (1.9549 × 10^7 Pa)
Volume of NH3 = (42.3 mol * 8.314 J/(mol·K) * 723.15 K) / (1.9549 × 10^7 Pa)