Iron pyrite (FeS2) is the form in which much of
the sulfur exists in coal. In the combustion of
coal, oxygen reacts with iron pyrite to produce
iron(III) oxide and sulfur dioxide, which is a
major source of air pollution and a substantial
contributor to acid rain. What mass of Fe2O3
is produced from the reaction is 77 L of oxygen
at 3.58 atm and 149â—¦C with an excess of iron
pyrite?
Answer in units of g.
If the sulfur dioxide that is generated above
is dissolved to form 5.5 L of aqueous solution,
what is the molar concentration of the
resulting sulfurous acid (H2SO3) solution?
Answer in units of M.
What mass of SO2 is produced in the burning
of 1 tonne (1 t = 1000 kg) of high-sulfur coal,
if the coal is 4% pyrite by mass?
Answer in units of kg.
From the previous problem, what is the volume
of the SO2 gas at 0.8 atm and 34â—¦C?
Answer in units of L.
To solve these questions, we need to use stoichiometry and the ideal gas law. Let's break down each question step by step:
1) What mass of Fe2O3 is produced from the reaction if 77 L of oxygen at 3.58 atm and 149°C are reacted with an excess of iron pyrite (FeS2)?
To solve this, we need to find the balanced chemical equation for the reaction between oxygen and iron pyrite. The chemical equation is:
4FeS2 + 11O2 -> 2Fe2O3 + 8SO2
From the balanced equation, we can see that 4 moles of FeS2 react with 11 moles of O2 to produce 2 moles of Fe2O3. Now, let's use the given information to calculate the number of moles of O2:
PV = nRT
P = 3.58 atm
V = 77 L
T = 149°C = 273 + 149 = 422 K (converted to Kelvin)
R = 0.0821 L·atm/(mol·K)
Rearranging the ideal gas law equation to solve for moles:
n = PV / RT = (3.58 atm) * (77 L) / (0.0821 L·atm/(mol·K) * 422 K) ≈ 12.59 moles
Now, using the stoichiometric ratio between O2 and Fe2O3, we can calculate the number of moles of Fe2O3:
(2 moles Fe2O3 / 11 moles O2) * 12.59 moles O2 = 2.29 moles Fe2O3
To convert moles of Fe2O3 to grams, we need to know the molar mass of Fe2O3, which is approximately 159.70 g/mol. Finally, we can calculate the mass of Fe2O3:
Mass = moles * molar mass = 2.29 moles * 159.70 g/mol ≈ 366.05 g
Therefore, the mass of Fe2O3 produced from the reaction is approximately 366.05 grams.
2) What is the molar concentration of the resulting sulfurous acid (H2SO3) solution when the generated sulfur dioxide (SO2) is dissolved to form 5.5 L of aqueous solution?
The balanced chemical equation for the conversion of SO2 to sulfurous acid (H2SO3) is:
SO2 + H2O -> H2SO3
The molar concentration (Molarity) is calculated by dividing the number of moles of solute (H2SO3) by the volume of the solution in liters (L). To find the molarity, we need to calculate the number of moles of H2SO3.
The molar ratio between SO2 and H2SO3 is 1:1. Therefore, the number of moles of H2SO3 will be equal to the number of moles of SO2.
Using the ideal gas law, we can calculate the number of moles of SO2:
PV = nRT
P = Pressure = ?
V = 5.5 L
R = 0.0821 L·atm/(mol·K)
T = 298 K (room temperature, 25°C)
However, the pressure (P) in the equation is not given. We need to know the pressure of the SO2 gas to calculate the number of moles. Please provide the pressure of SO2 gas or any additional information needed for this calculation.
3) What mass of SO2 is produced in the burning of 1 tonne (1 t = 1000 kg) of high-sulfur coal, if the coal is 4% pyrite by mass?
To solve this question, we need to calculate the amount of pyrite in the coal and then determine the amount of SO2 produced.
Given:
Mass of coal = 1 tonne = 1000 kg
Pyrite percentage in coal = 4%
First, find the mass of pyrite in the coal:
Mass of pyrite = Mass of coal * Pyrite percentage
= 1000 kg * 0.04
= 40 kg
Next, we need to find the number of moles of pyrite:
Molar mass of pyrite (FeS2) = (1 x atomic mass of Fe) + (2 x atomic mass of S)
= (1 x 55.845 g/mol) + (2 x 32.06 g/mol)
≈ 119.99 g/mol
Number of moles of pyrite = mass of pyrite / molar mass of pyrite
= 40 kg / (119.99 g/mol)
≈ 333.36 moles
From the balanced chemical equation in the previous question (4FeS2 + 11O2 -> 2Fe2O3 + 8SO2), we know that 4 moles of FeS2 produce 8 moles of SO2. Therefore, 333.36 moles of FeS2 will produce:
333.36 moles FeS2 * (8 moles SO2 / 4 moles FeS2) = 666.72 moles SO2
Finally, we convert moles of SO2 to grams using the molar mass of SO2:
Molar mass of SO2 = (1 x atomic mass of S) + (2 x atomic mass of O)
= (1 x 32.06 g/mol) + (2 x 16.00 g/mol)
≈ 64.06 g/mol
Mass of SO2 = moles of SO2 * molar mass of SO2
= 666.72 moles * 64.06 g/mol
≈ 42,700.03 g
Therefore, the mass of SO2 produced in the burning of 1 tonne of high-sulfur coal is approximately 42,700.03 grams.
4) What is the volume of the SO2 gas at 0.8 atm and 34°C, considering the conditions provided?
To solve this, we need to use the ideal gas law again:
PV = nRT
P = 0.8 atm
V = ?
R = 0.0821 L·atm/(mol·K)
T = 34°C = 273 + 34 = 307 K (converted to Kelvin)
Rearranging the ideal gas law equation to solve for volume:
V = (nRT) / P
We already calculated the number of moles of SO2 in the previous question, which is 666.72 moles. Now we can substitute this value into the equation:
V = (666.72 moles) * (0.0821 L·atm/(mol·K)) * (307 K) / (0.8 atm)
≈ 20,165.45 L
Therefore, the volume of SO2 gas at 0.8 atm and 34°C is approximately 20,165.45 liters.