If a sulfuric acid plant is to produce 10 ton/day of 98% sulfuric acid, How much sulfur is needed, how much air is required and how much water is used?
To calculate the amount of sulfur, air, and water required to produce 10 tons/day of 98% sulfuric acid, we need to consider the chemical equation and stoichiometry of the reaction.
The balanced chemical equation for the production of sulfuric acid can be written as:
S (sulfur) + O2 (air) + H2O (water) → H2SO4 (sulfuric acid)
From the equation, we can see that one mole of sulfur reacts with one mole of oxygen and two moles of water to produce two moles of sulfuric acid.
Step 1: Calculate the number of moles of sulfuric acid produced per day.
To get the number of moles, we need to convert the given weight of sulfuric acid to moles using its molar mass.
Molar mass of sulfuric acid (H2SO4):
(2 * 1.008) + (1.008 * 2) + (16.00 * 4) = 98.08 g/mol
Moles of sulfuric acid produced per day:
(10,000 kg/day * 1000 g/kg) / 98.08 g/mol = 101.9 mol/day
Step 2: Determine the moles of sulfur required.
From the balanced equation, we can see that the molar ratio between sulfur and sulfuric acid is 1:2. Therefore, we need half the moles of sulfuric acid to calculate the moles of sulfur.
Moles of sulfur required per day:
101.9 mol/day / 2 = 50.95 mol/day
Step 3: Calculate the mass of sulfur required.
Using the molar mass of sulfur, we can convert the moles of sulfur to grams.
Molar mass of sulfur (S):
32.06 g/mol
Mass of sulfur required per day:
50.95 mol/day * 32.06 g/mol = 1633.19 g/day = 1.63 kg/day
So, to produce 10 tons/day of 98% sulfuric acid, approximately 1.63 kg/day of sulfur is required.
Step 4: Determine the moles of air required.
Again, referring to the balanced equation, we can see that the molar ratio between oxygen (from air) and sulfuric acid is 1:1. Therefore, we need the same number of moles of air as moles of sulfuric acid produced.
Moles of air required per day:
101.9 mol/day
Step 5: Calculate the volume of air required.
To calculate the volume of air required, we need to consider the molar volume of ideal gas at standard conditions (STP), which is approximately 22.4 L/mol.
Volume of air required per day:
101.9 mol/day * 22.4 L/mol = 2281.36 L/day
So, approximately 2281.36 liters of air are required to produce 10 tons/day of 98% sulfuric acid.
Step 6: Determine the moles of water required.
Again, referring to the balanced equation, we can see that the molar ratio between water and sulfuric acid is 2:1. Therefore, we need twice the moles of water to calculate the moles of sulfuric acid.
Moles of water required per day:
2 * 101.9 mol/day = 203.8 mol/day
Step 7: Calculate the mass of water required.
Using the molar mass of water, we can convert the moles of water to grams.
Molar mass of water (H2O):
(1.008 * 2) + (16.00) = 18.02 g/mol
Mass of water required per day:
203.8 mol/day * 18.02 g/mol = 3673.4 g/day = 3.67 kg/day
So, to produce 10 tons/day of 98% sulfuric acid, approximately 3.67 kg/day of water is required.
In summary:
- Sulfur required: 1.63 kg/day
- Air required: 2281.36 liters/day
- Water used: 3.67 kg/day