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 70 L of oxygen at 4.12 atm
and 167�C with an excess of iron pyrite?
Answer in units of g

figure the moles of O with ideal gas law.

n=PV/RT

take 1/3 of that, that then is the moles of Fe2O3
calculate the mass of that.

02=31.998. PV=NIGT

To determine the mass of Fe2O3 produced, we can follow these steps:

1. Convert the given volume of oxygen to moles using the ideal gas law.
2. Use the stoichiometry of the reaction to determine the moles of Fe2O3 produced.
3. Convert moles of Fe2O3 to grams using the molar mass of Fe2O3.

Let's begin by calculating the moles of oxygen:

1. Convert the given temperature to Kelvin:
T(K) = 167°C + 273.15 = 440.15 K

2. Convert the given pressure to atm:
P(atm) = 4.12 atm

3. Use the ideal gas law to calculate the moles of oxygen:
PV = nRT
(4.12 atm)(70 L) = n(0.0821 atm·L/mol·K)(440.15 K)
n = (4.12 atm * 70 L) / (0.0821 atm·L/mol·K * 440.15 K)
n ≈ 7.02 mol

Now, we can determine the moles of Fe2O3 produced using the stoichiometry of the reaction:

The balanced equation for the combustion of iron pyrite is:
4 FeS2 + 11 O2 -> 2 Fe2O3 + 8 SO2

From the equation, we can see that 4 moles of FeS2 produce 2 moles of Fe2O3. Therefore, the moles of Fe2O3 produced can be calculated as follows:

4 moles FeS2 = 2 moles Fe2O3
x moles FeS2 = 7.02 mol
x = (7.02 mol * 2 mol Fe2O3) / 4 mol FeS2
x = 3.51 mol Fe2O3

Finally, we can convert moles of Fe2O3 to grams by multiplying by the molar mass of Fe2O3, which is approximately 159.7 g/mol:

Mass of Fe2O3 = 3.51 mol * 159.7 g/mol
Mass of Fe2O3 ≈ 559.2 g (to three significant figures)

Therefore, approximately 559.2 grams of Fe2O3 will be produced.

To determine the mass of Fe2O3 produced, we need to do the following steps:

1. Find the moles of oxygen gas (O2) using the ideal gas law.
2. Set up the balanced chemical equation to find the mole ratio of oxygen to Fe2O3.
3. Use the mole ratio to calculate the moles of Fe2O3 produced.
4. Convert moles of Fe2O3 to grams.

Step 1: Find moles of oxygen gas (O2)
The ideal gas law equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.

Given:
Pressure (P) = 4.12 atm
Volume (V) = 70 L
Temperature (T) = 167 ºC = 167 + 273 = 440 K

Rearranging the ideal gas law equation to solve for moles:
n = PV / RT

n = (4.12 atm)(70 L) / (0.0821 L·atm/mol·K)(440 K)
n ≈ 9.39 mol

Step 2: Balanced chemical equation
The balanced chemical equation for the reaction is:
4FeS2 + 11O2 → 2Fe2O3 + 8SO2

From this equation, we can see that 11 moles of oxygen (O2) react to produce 2 moles of iron(III) oxide (Fe2O3).

Step 3: Calculate moles of Fe2O3 produced
Using the mole ratio from the balanced chemical equation, we can calculate the moles of Fe2O3 produced.

moles of Fe2O3 = (9.39 mol O2) × (2 mol Fe2O3 / 11 mol O2)
moles of Fe2O3 ≈ 1.71 mol

Step 4: Convert moles of Fe2O3 to grams
The molar mass of Fe2O3 can be found by summing the atomic masses of its atoms:
2(Fe) + 3(O) = (2 * 55.85 g/mol) + (3 * 16.00 g/mol) = 159.69 g/mol

mass of Fe2O3 = (1.71 mol Fe2O3) × (159.69 g/mol Fe2O3)
mass of Fe2O3 ≈ 273.29 g

Therefore, the mass of Fe2O3 produced from 70 L of oxygen at 4.12 atm and 167 ºC with an excess of iron pyrite is approximately 273.29 grams.