modified Haber process for making ammonia [ H2 + N2 --> NH3 ] is conducted at 550.ºC and 2.50x102 atm. If 7.55 kg of nitrogen (the limiting reactant) is used and the process goes to completion, what volume of ammonia is produced?
Answer
L
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how many moles N2 is that?
how many moles NH3 are produced?
PV=kT
and V=22.4 at STP
To find the volume of ammonia produced, we need to 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 in Kelvin.
First, let's calculate the number of moles of nitrogen (N2) used:
Given mass of nitrogen = 7.55 kg
The molar mass of N2 = 28 g/mol
Therefore, the number of moles of nitrogen = mass / molar mass
= 7550 g / 28 g/mol
= 269.64 mol
According to the balanced equation, the molar ratio of nitrogen (N2) to ammonia (NH3) is 1:2. So, the number of moles of ammonia produced is double the number of moles of nitrogen used.
Number of moles of ammonia = 2 * (269.64 mol)
= 539.28 mol
Now, let's convert the temperature from Celsius to Kelvin:
Given temperature = 550 ºC
Temperature in Kelvin = Celsius + 273.15
= 550 ºC + 273.15
= 823.15 K
Next, we need to calculate the volume of ammonia produced by rearranging the ideal gas law equation:
V = (n * R * T) / P
Let's substitute the values into the equation:
n (number of moles) = 539.28 mol
R (ideal gas constant) = 0.0821 L·atm/mol·K (at 1 atm pressure)
T (temperature in Kelvin) = 823.15 K
P (pressure) = 2.50 x 10^2 atm
V = (539.28 mol * 0.0821 L·atm/mol·K * 823.15 K) / (2.50 x 10^2 atm)
V ≈ 14837.55 L
Therefore, approximately 14837.55 liters of ammonia will be produced in the Haber process.
To find the volume of ammonia produced, we need to use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
First, let's find the number of moles of nitrogen (N2) used in the reaction. We know that the molecular weight of nitrogen is 28 g/mol. Therefore, the number of moles of nitrogen used can be calculated as follows:
Number of moles of N2 = mass of nitrogen / molar mass of nitrogen
= 7550 g / 28 g/mol
= 269.64 mol
Since the reaction goes to completion, all the nitrogen would be converted into ammonia (NH3) in a 1:1 stoichiometric ratio. Therefore, the number of moles of ammonia produced would also be 269.64 mol.
Now, let's use the ideal gas law equation to find the volume of ammonia:
PV = nRT
Since we are looking for the volume, we can rearrange the equation to solve for V:
V = (nRT) / P
Substituting the values we have:
V = (269.64 mol * 0.0821 atm L/mol K * 550 °C) / (2.50 x 10^2 atm)
Note: Make sure to convert the temperature from Celsius to Kelvin by adding 273.15.
V = (269.64 mol * 0.0821 atm L/mol K * (550 + 273.15) K) / (2.50 x 10^2 atm)
Now, we can calculate the volume of ammonia:
V = (269.64 * 0.0821 * 823.15) / (2.50 x 10^2)
= 19.346 L
Therefore, the volume of ammonia produced is approximately 19.346 L.