Nitric acid can be produced from NH3 in three steps process:
I)4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
II)2NO(g) + O2(g) → 2NO2(g)
III)3NO2(g) + H2O(l) →2HNO3(aq) + NO(g)
% yield of Ist, IInd & IIIrd are respectively 50 % , 60 % & 80 % respectively the what volume of NH3(g) at STP required to produced 1575 kg of HNO3.
3500 Litre
To calculate the volume of NH3(g) required to produce 1575 kg of HNO3, we need to determine the limiting reactant and use stoichiometry.
First, let's calculate the moles of HNO3 produced:
Molar mass of HNO3 = 1 + 14 + 16(3) = 63 g/mol
Mass of HNO3 = 1575 kg = 1575000 g
Moles of HNO3 = mass of HNO3 / molar mass = 1575000 g / 63 g/mol = 25000 mol
Now, let's work through the steps of the process and determine the limiting reactant at each step:
Step I:
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
The percent yield for Step I is given as 50%.
Since we know that 4 moles of NH3 are required per mole of NO, we can write the following proportion:
4 moles of NH3 : 4 moles of NO :: x moles of NH3 : 25000 moles of HNO3
x = (4 moles of NH3 * 25000 moles of HNO3) / 4 moles of NO
x = 25000 moles of NH3
Since the percent yield is 50%, the actual moles of NH3 produced will be:
Moles of NH3 produced in Step I = (50/100) * 25000 moles = 12500 moles
Step II:
2NO(g) + O2(g) → 2NO2(g)
The percent yield for Step II is given as 60%.
Since we know that 2 moles of NO are required per mole of NO2, we can write the following proportion:
2 moles of NO : 2 moles of NO2 :: 12500 moles of NO : x moles of NO2
x = (2 moles of NO2 * 12500 moles of NO) / 2 moles of NO
x = 12500 moles of NO2
Since the percent yield is 60%, the actual moles of NO2 produced will be:
Moles of NO2 produced in Step II = (60/100) * 12500 moles = 7500 moles
Step III:
3NO2(g) + H2O(l) → 2HNO3(aq) + NO(g)
The percent yield for Step III is given as 80%.
Since we know that 3 moles of NO2 are required per mole of HNO3, we can write the following proportion:
3 moles of NO2 : 2 moles of HNO3 :: 7500 moles of NO2 : x moles of HNO3
x = (2 moles of HNO3 * 7500 moles of NO2) / 3 moles of NO2
x = 5000 moles of HNO3
Now, we can calculate the moles of NH3 required for 5000 moles of HNO3, using the stoichiometric ratio from Step I:
4 moles of NH3 : 4 moles of NO :: y moles of NH3 : 5000 moles of HNO3
y = (4 moles of NH3 * 5000 moles of HNO3) / 4 moles of NO
y = 5000 moles of NH3
Finally, we can calculate the volume of NH3 at STP required for 5000 moles:
1 mole of NH3 occupies 22.4 L at STP,
Therefore, Volume of NH3 required = 5000 moles * 22.4 L/mol = 112000 L
Therefore, the volume of NH3(g) at STP required to produce 1575 kg of HNO3 is 112000 L.
To find the volume of NH3(g) required to produce 1575 kg of HNO3, we need to work backward through the given reactions and calculate the amount of NH3 needed at each step.
Let's start by calculating the moles of HNO3 produced:
Given mass of HNO3 = 1575 kg
Molar mass of HNO3 = 63 g/mol
Moles of HNO3 = (1575 kg / 63 g/mol) = 25,000 mol
Now, let's calculate the moles of NO2 needed in the third step:
Mole ratio of HNO3 to NO2 (from the balanced equation III) = 2:3
Moles of NO2 = (3/2) * Moles of HNO3 = (3/2) * 25,000 mol = 37,500 mol
Since the % yield of the third step is given as 80%, we can calculate the actual moles of NO2 produced:
Actual moles of NO2 = (80/100) * Moles of NO2 = (80/100) * 37,500 mol = 30,000 mol
Next, let's calculate the moles of NO needed in the first step:
Mole ratio of NO2 to NO (from the balanced equation I) = 4:2
Moles of NO = (2/4) * Moles of NO2 = (2/4) * 30,000 mol = 15,000 mol
Since the % yield of the first step is given as 50%, we can calculate the actual moles of NO produced:
Actual moles of NO = (50/100) * Moles of NO = (50/100) * 15,000 mol = 7,500 mol
Finally, let's calculate the volume of NH3(g) at STP (Standard Temperature and Pressure) required to produce 7,500 moles:
Molar volume of any ideal gas at STP = 22.4 L/mol
Volume of NH3(g) = Moles of NH3 * Molar volume at STP
Volume of NH3(g) = 7,500 mol * 22.4 L/mol = 168,000 L
Therefore, the volume of NH3(g) at STP required to produce 1575 kg of HNO3 is 168,000 liters.
1575 kg
1575000g
1575000/17 = estimated 9E4 mol
All of the NH3 at the beginning produces HNO3 (at 100% yield) so 9E4 mol NH3 will produce 9E4 mol HNO3. That will be 9E4 mols x 22.4 mol/L = estd 2E6 L if everything is 100%. But these are not. The efficiency is 0.5 x 0.6 x 0.8 = 0.24 so
2E6L at 100%/0.24 = ? volume needed at 24%. Note: That 2E6 L is an estimate. You need to sharpen that number.