If 3.20 kg of Sulphur reacts with oxygen, calculate the volume in millileters of sulphur dioxide gas formed at 30.5 degrees Celsius and 1.04 atm.

S + O2 --> SO2

3.20 kg = 3200 grams. Convert to mols. mols = grams/molar mass.

Use the coefficients in the balanced equation to convert mols S to mols SO2

Now use PV = nRT, substitute mols for n, substitute the conditions in the problem (P and T) and solve for volume. Don't forget T must be in kelvin.

To calculate the volume of sulfur dioxide gas formed, we can use the ideal gas law equation: PV = nRT, where P represents the pressure, V represents the volume, n represents the number of moles of gas, R is the ideal gas constant, and T represents the temperature in Kelvin.

First, we need to determine the number of moles of sulfur dioxide gas formed by using the balanced chemical equation for the reaction:

S + O2 -> SO2

From the balanced equation, we can see that 1 mole of sulfur reacts with 1 mole of oxygen to produce 1 mole of sulfur dioxide. Therefore, the number of moles of sulfur dioxide formed will be the same as the number of moles of sulfur used.

Given that we have 3.20 kg of sulfur, we need to convert the mass to moles by using the molar mass of sulfur, which is approximately 32.06 g/mol. Converting 3.20 kg to grams, we have:

3.20 kg = 3200 g

Now, we can calculate the number of moles of sulfur:

moles of sulfur = mass of sulfur / molar mass of sulfur
= 3200 g / 32.06 g/mol
= 99.88 mol (rounded to two decimal places)

Since we have established that the number of moles of sulfur dioxide formed is equal to the number of moles of sulfur used, we have 99.88 moles of sulfur dioxide gas formed.

Now we can solve for the volume of sulfur dioxide gas using the ideal gas law equation:

PV = nRT

To calculate the volume, we need to convert temperature from degrees Celsius to Kelvin:

T(K) = T(°C) + 273.15
= 30.5°C + 273.15
= 303.65 K

Also, we know the pressure is 1.04 atm.

Now solving for volume:

V = (nRT) / P
= (99.88 mol) * (0.0821 atm·L/mol·K) * (303.65 K) / (1.04 atm)
= 2391.86 L (rounded to two decimal places)

However, the question asks for the volume in milliliters (ml), so we need to convert liters to milliliters:

1 L = 1000 mL

Therefore, the volume of sulfur dioxide gas formed is 2391.86 L * 1000 mL/L = 2391860 mL (rounded to two decimal places).

So, the volume of sulfur dioxide gas formed at 30.5 degrees Celsius and 1.04 atm is approximately 2391860 mL.