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 71 L of oxygen at 3.72 atm and 158 degrees C with an excess of iron pyrite?
Answer in units of g

Deez nutz in your mouth

To find the mass of Fe2O3 produced from the given amount of oxygen, we first need to determine the moles of oxygen gas (O2) using the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)

Given:
Pressure (P) = 3.72 atm
Volume (V) = 71 L
Temperature (T) = 158°C = 158 + 273 = 431 K

Plugging these values into the ideal gas law equation:

n = PV / RT

n = (3.72 atm * 71 L) / (0.0821 L.atm/mol.K * 431 K)
n = 0.1267 mol

Since oxygen reacts with iron pyrite in a 1:1 ratio, we can say that the moles of iron(III) oxide (Fe2O3) produced will also be 0.1267 mol.

Next, we need to calculate the molar mass of Fe2O3, which is the sum of the atomic masses of iron (Fe) and oxygen (O). The atomic mass of Fe is 55.85 g/mol, and the atomic mass of O is 16.00 g/mol.

Molar mass of Fe2O3 = (2 x atomic mass of Fe) + (3 x atomic mass of O)
Molar mass of Fe2O3 = (2 x 55.85 g/mol) + (3 x 16.00 g/mol)
Molar mass of Fe2O3 = 159.70 g/mol

To find the mass of Fe2O3 produced:
Mass = moles * molar mass
Mass = 0.1267 mol * 159.70 g/mol
Mass ≈ 20.228 g

Therefore, approximately 20.228 grams of Fe2O3 is produced from 71 liters of oxygen at 3.72 atm and 158 degrees Celsius with an excess of iron pyrite.

To find the mass of Fe2O3 produced, we first need to determine the balanced chemical equation for the reaction:

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.

Given:
Volume of oxygen (O2) = 71 L
Pressure of oxygen (O2) = 3.72 atm
Temperature of oxygen (O2) = 158 °C = 158 + 273 = 431 K

To use the Ideal Gas Law (PV = nRT) to find the number of moles of oxygen, we need to convert the temperature to Kelvin (K):
T = 431 K

Next, we can use the ideal gas equation to calculate the number of moles of oxygen (O2):

PV = nRT

n = (PV) / (RT)
= (3.72 atm x 71 L) / (0.0821 L∙atm/mol∙K x 431 K)
≈ 1.45 moles of O2

From the balanced equation, we know that 4 moles of FeS2 will produce 2 moles of Fe2O3. This means that every 2 moles of Fe2O3 correspond to 4 moles of FeS2.

Therefore, if we have 1.45 moles of O2, we would need half that amount (0.725 moles) of FeS2.

Now we can use the molar mass of Fe2O3 to determine the mass of Fe2O3 produced:

Molar mass of Fe2O3 = (2 x atomic mass of Fe) + (3 x atomic mass of O)
= (2 x 55.845 g/mol) + (3 x 16.00 g/mol)
= 111.69 g/mol

Mass of Fe2O3 = moles of Fe2O3 x molar mass of Fe2O3
= 0.725 moles x 111.69 g/mol
≈ 81.01 g

Therefore, approximately 81.01 grams of Fe2O3 will be produced.

Here is the formula for solving theser stoichiometry problems.

1. Write the equation and balance it.
2. Convert what you have into mols. mols = grams/molar mass.
3. Using the coefficients in the balanced equation, convert mols of what you have (from step 2) to mols of the product.
4. Now convert mols of the product to grams. g = mols x molar mass.