Sulfuric acid is produced commercially by the reactions indicated below at the appropriate percent yields for each step. The second step utilizes air as a source of oxygen and heats it to 4000C in the presence of vanadium (V) oxide, a catalyst. Assuming air is 21% oxygen by volume, what volume of air at 190C and 754 torr is required to produce enough SO3 to eventually yield 1.00 metric ton (1000kg=1metric ton) of sulfuric acid?
S(s) + O2(g)--->SO2(g) 98.3%
2SO2(g) + O2(g)--->2SO3(g) 94.5%
H2SO4(l) + SO3(g) --->H2S2O7(s) 89.7%
H2S2O7(s) + H2O(l)--->2H2SO4(l) 91.2%
To solve this problem, we need to calculate the volume of air required to produce enough SO3 to eventually yield 1.00 metric ton of sulfuric acid.
Step 1: Calculate the moles of SO3 required.
Given that 1 metric ton is equal to 1000 kg, we need to find the moles of SO3 in 1000 kg of sulfuric acid.
The molar mass of SO3 is:
Sulfur (S): 32.07 g/mol
Oxygen (O): 16.00 g/mol * 3 = 48.00 g/mol
Therefore, the molar mass of SO3 is 32.07 g/mol + 48.00 g/mol = 80.07 g/mol.
To convert 1000 kg to grams, we multiply by 1000:
1000 kg * 1000 g/kg = 1,000,000 g
Now, divide the mass of SO3 (1,000,000 g) by the molar mass of SO3 (80.07 g/mol) to find the moles of SO3 required:
Moles of SO3 = 1,000,000 g / 80.07 g/mol = 12,491.47 mol
Step 2: Calculate the moles of O2 required for the reaction.
Since the reaction S(s) + O2(g) ---> SO2(g) has a percent yield of 98.3% and 1 mol of SO3 is produced from 2 mol of SO2, we need to calculate the moles of O2 required based on the percent yield and stoichiometry.
From the reaction stoichiometry, we know that 2 mol of SO2 react with 1 mol of O2 to produce 2 mol of SO3.
Using the percent yield of 98.3%, we can calculate the moles of O2 required:
Moles of O2 = (12,491.47 mol SO3 / 2) * (1 mol O2 / 98.3%) = 6369.14 mol O2
Step 3: Calculate the volume of air required.
To calculate the volume of air required, we need to convert the moles of O2 to moles of air. Since air is 21% oxygen, the remaining 79% is nitrogen.
Moles of nitrogen (N2) = Moles of O2 / 0.21
Moles of nitrogen = 6369.14 mol O2 / 0.21 = 30,328.29 mol N2
To convert moles of nitrogen to volume, we can use the ideal gas law: PV = nRT
Assuming the temperature is 190C = 463K and the pressure is 754 torr = 754/760 atm, we can rearrange the ideal gas law equation to solve for volume (V):
V = (nRT) / P
Using the values:
n = 30,328.29 mol N2
R = 0.0821 (atm * L) / (mol * K)
T = 463K
P = 754/760 atm
V = (30,328.29 mol N2 * 0.0821 (atm * L) / (mol * K) * 463K) / (754/760 atm)
V = 37,079.30 L
Therefore, the volume of air required to produce enough SO3 to eventually yield 1.00 metric ton of sulfuric acid is 37,079.30 liters.