What volume of air containing 21% of oxygen (by volume) is required to completely burn 10g of sulphur which has a purity level of 98%.
lots of words, but basically devoid of content
oobleck is right on. I think the questions (most of them anyway) are made up too.
To determine the volume of air required to completely burn sulfur, we first need to balance the chemical equation for the combustion of sulfur.
The chemical equation for the combustion of sulfur is:
S(s) + O2(g) -> SO2(g)
From the balanced equation, we can see that one mole of sulfur requires one mole of O2 gas to generate one mole of SO2 gas. Therefore, we need to convert the given mass of sulfur to moles using its molar mass and then use stoichiometry to determine the number of moles of O2 required.
Step 1: Calculate the moles of sulfur (S)
Given:
Mass of Sulfur (S) = 10g
Purity level of sulfur = 98%
To determine the number of moles of sulfur, we first need to calculate the molar mass of sulfur using the periodic table.
The atomic mass of sulfur (S) is approximately 32 g/mol.
Mass of Sulfur (S) = Number of moles of Sulfur (S) × Molar mass of Sulfur (S)
Number of moles of Sulfur (S) = Mass of Sulfur (S) / Molar mass of Sulfur (S)
= 10g / (32g/mol)
≈ 0.3125 mol
Since the purity level of sulfur is given, we need to adjust the number of moles of sulfur accordingly:
Adjusted moles of Sulfur (S) = Number of moles of Sulfur (S) × (Purity level / 100)
= 0.3125 mol × (98/100)
≈ 0.306 mol
Step 2: Calculate the moles of oxygen (O2)
From the balanced chemical equation, we know that one mole of sulfur reacts with one mole of oxygen (O2). Therefore, the number of moles of oxygen required is the same as the number of moles of sulfur.
Moles of Oxygen (O2) = Adjusted moles of Sulfur (S) = 0.306 mol
Step 3: Calculate the volume of air
Given that the air contains 21% oxygen by volume, we can calculate the volume of air using the ratio of oxygen to air in the reaction.
Volume of Air = Moles of Oxygen (O2) / Fraction of Oxygen in Air
Fraction of Oxygen in Air = 21% = 0.21
Volume of Air = 0.306 mol / 0.21
≈ 1.457 mol
Since 1 mole of a gas at standard temperature and pressure (STP) occupies 22.4 liters, we can convert the moles of air to volume in liters.
Volume of Air = 1.457 mol × 22.4 L/mol
≈ 32.575 L
Therefore, approximately 32.575 liters of air containing 21% oxygen by volume would be required to completely burn 10g of sulfur with a purity level of 98%.