Sulfur trioxide, SO3, is produced in enormous quantities each year for use in the synthesis of sulfuric acid.

S(s) + O2(g) -> SO2(g)
2SO2(g) + O2(g) -> 2SO3(g)

What volume of O2 (g) at 350 degrees Celsius and a pressure of 5.25 atm is needed to completely convert 5.00g of sulfur to sulfur trioxide?

Okay, I know this is 10 years old. I just found it because I had this exact same problem on a homework assignment. Here's what I got after a bit of work:

V = Volume of O2

350.°C -> 623 K

5.00 g S / 32.06 g S = 0.156 mol S

0.156 / 2 = 0.078 mol O2

0.156 + 0.078 = 0.234 mol O2

V = (0.234 * 0.08206 * 623) / 5.25 = 2.28 L O2

I'm pretty sure this is right. I hope it helps whoever comes next!

The long way, but perhaps easier to understand, is to convert 5 g S to moles, then to mole O2 needed and moles SO2 produced in the first equation. Take the mole SO2 produced in equation 1 and use that to determine moles oxygen from equation 2. Add the two to find total moles O2. Then use PV = nRT to find volume O2. Check my thinking.

I forgot to include the equations I used.

I used this one:
PV = nRT

But we were solving for volume, so I changed it to this:
V = nRT / P

Well, looks like we have some sulfur party going on here! Let's do some chemistry calculations and clown around with the numbers!

First, we need to find the number of moles of sulfur (S) using its molar mass. The molar mass of sulfur is approximately 32.06 g/mol. So, 5.00 grams of sulfur is equal to 5.00 g / 32.06 g/mol = 0.156 mol of Sulfur.

According to the balanced equation, we need 1 mol of sulfur to produce 1 mol of sulfur dioxide (SO2) and 1 mol of sulfur trioxide (SO3). So, we'll end up with 0.156 mol of SO2 and 0.156 mol of SO3.

Now, let's focus on finding the volume of oxygen gas (O2) at 350 degrees Celsius and 5.25 atm needed for the conversion.

We can use the ideal gas law, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

First, we need to convert Celsius to Kelvin. So, 350 degrees Celsius + 273.15 = 623.15 Kelvin.

Now, we can rearrange the equation to solve for volume:
V = nRT / P

Plugging in the values:
V = (0.156 mol) * (0.0821 L·atm/mol·K) * (623.15 K) / (5.25 atm)

Calculating this, we get V ≈ 0.146 L (rounded to 3 significant figures).

So, we need approximately 0.146 liters of oxygen gas to convert 5.00 grams of sulfur to sulfur trioxide! Hope that puts a smile on your face!

To solve this problem, we need to use the given balanced equation and stoichiometry to determine the volume of O2 gas required.

First, let's convert the mass of sulfur (S) to moles using its molar mass.
The molar mass of S = 32.07 g/mol.

So, moles of Sulfur (S) = mass of Sulfur (g) / molar mass of Sulfur (g/mol)
Moles of Sulfur (S) = 5.00 g / 32.07 g/mol
Moles of Sulfur (S) = 0.156 mol

According to the balanced equation, the mole ratio between S and O2 is 1:1. This means that for every one mole of sulfur, we need one mole of O2.

Now we need to find the moles of O2 required. Since the mole ratio is 1:1, the moles of O2 required will be the same as the moles of sulfur.

Moles of O2 required = 0.156 mol

Next, we'll need to use the ideal gas law equation to find the volume of O2 gas at the given temperature and pressure.

The ideal gas law equation is:
PV = nRT

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

First, let's convert the temperature from Celsius to Kelvin:
T(K) = 350°C + 273.15
T(K) = 623.15 K

Now, we can rearrange the ideal gas law equation to solve for the volume:
V = (nRT) / P

Substituting the values into the equation, we get:
V = (0.156 mol * 0.0821 L·atm/mol·K * 623.15 K) / 5.25 atm

Calculating this, we find:
V ≈ 1.11 L (rounded to two decimal places)

Therefore, approximately 1.11 liters of O2 gas at 350°C and a pressure of 5.25 atm are needed to completely convert 5.00g of sulfur to sulfur trioxide.