Iron pyrite (FeS2) is the form in whichmuch 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 87 L of oxygen at 2.38 atm
and 163�C with an excess of iron pyrite?
4FeS2 + 11O2 ==> 8SO2 + 2Fe2O3
Use PV = nRT and solve for n = number mols O2.
Using the coefficients in the balanced equation, convert mols O2 to mols Fe2O3.
Then convert mols Fe2O3 to grams. g = mols x molar mass.
To find the mass of Fe2O3 produced, we need to use the ideal gas law to determine the number of moles of oxygen (O2).
Given:
Volume of oxygen (V) = 87 L
Pressure of oxygen (P) = 2.38 atm
Temperature of oxygen (T) = 163°C = 163 + 273 = 436 K
Molecular weight of oxygen (O2) = 32 g/mol
First, let's convert the temperature from Celsius to Kelvin:
T = 163 + 273 = 436 K
Next, we can use the ideal gas law equation to find the number of moles of oxygen:
PV = nRT
R is the ideal gas constant, which is 0.0821 L·atm/(mol·K).
n = (PV) / (RT)
n = (2.38 atm * 87 L) / (0.0821 L·atm/(mol·K) * 436 K)
n = 2.606 mol
According to the balanced equation for the reaction of iron pyrite with oxygen, 4 moles of iron pyrite react with 11 moles of oxygen to produce 4 moles of Fe2O3.
Since the iron pyrite is in excess, we can assume that all the oxygen reacts.
From the stoichiometry, we can determine the number of moles of Fe2O3 produced:
4 moles of iron pyrite react with 11 moles of oxygen to produce 4 moles of Fe2O3
So, 11 moles of oxygen produce 4 moles of Fe2O3
Using the mole ratio, we can find the number of moles of Fe2O3 produced:
n(Fe2O3) = (4 mol Fe2O3 / 11 mol O2) * n(O2)
n(Fe2O3) = (4 mol / 11 mol) * 2.606 mol
n(Fe2O3) = 0.947 mol
Finally, we can calculate the mass of Fe2O3 produced using the molar mass of Fe2O3:
Molar mass of Fe2O3 = (2 * molar mass of Fe) + (3 * molar mass of O)
Molar mass of Fe2O3 = (2 * 55.845 g/mol) + (3 * 16.00 g/mol)
Molar mass of Fe2O3 = 159.69 g/mol
Mass of Fe2O3 = n(Fe2O3) * molar mass of Fe2O3
Mass of Fe2O3 = 0.947 mol * 159.69 g/mol
Mass of Fe2O3 = 151.27 g
Therefore, approximately 151.27 grams of Fe2O3 are produced.
To find the mass of Fe2O3 produced, we can follow these steps:
Step 1: Calculate the number of moles of oxygen gas (O2) using the Ideal Gas Law equation.
PV = nRT
Where:
P = pressure (2.38 atm)
V = volume (87 L)
n = number of moles of gas (to be calculated)
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (163°C = 436 K)
Rearranging the equation and solving for n:
n = PV / RT
Step 2: Convert the number of moles of O2 to the number of moles of Fe2O3 using the balanced chemical equation.
The balanced chemical equation for the combustion of iron pyrite with oxygen is:
4FeS2 + 11O2 -> 2Fe2O3 + 8SO2
From the equation, we can see that 11 moles of O2 react to produce 2 moles of Fe2O3.
So, the number of moles of Fe2O3 = (n of O2) x (2 mol Fe2O3 / 11 mol O2)
Step 3: Calculate the mass of Fe2O3 using the molar mass of Fe2O3.
The molar mass of Fe2O3 is:
2(55.85 g/mol of Fe) + 3(16.00 g/mol of O) = 159.70 g/mol of Fe2O3
So, the mass of Fe2O3 = (number of moles of Fe2O3) x (molar mass of Fe2O3)
Let's calculate the values now:
Step 1:
n = (2.38 atm) x (87 L) / (0.0821 L·atm/(mol·K) x 436 K)
Step 2:
Number of moles of Fe2O3 = (n of O2) x (2 mol Fe2O3 / 11 mol O2)
Step 3:
Mass of Fe2O3 = (number of moles of Fe2O3) x (molar mass of Fe2O3)
By plugging in the values and performing the calculations, you'll find the mass of Fe2O3 produced from 87 L of oxygen at 2.38 atm and 163°C with an excess of iron pyrite.