*Only need input not doing it for me. PLEASE
QUESTION: Some hydrocarbon fuels have sulfur present and contribute to air pollution. When this solid sulfur is burned in the presence of oxygen, the noxious, toxic gas sulfur dioxide results. If there is .50 grams of sulfur and 1.50 grams of oxygen in each tank of gasoline, calculate the volume of sulfur dioxide gas that will be expelled into the atmosphere per each tank of gasoline at STP.
Ok so using STP I would use values of R=.0821 P=1 atm and T=273.15K???
Is the formula that I would use be PV=nRT????
I would covert Grams into moles using atomic weight?
Would I use the atomic weight of one O molecule or do I use the atomic weight of two O molecules?
When I came up with the calculations of my formula, is then the answer in atm?
It does not matter how much oxygen is in the tank, the SO2 is formed from S in the tank and O in the atmosphere.
Each mole of S (32 g) will form 1 mole of SO2, with a mass of 64 g. Thus one g of SO2 will be formed from 0.5 g of S. That is 1/64 mole of SO2.
Since each mole at STP occupies 22.4 l, you will make 22.4/64 = 0.35 liters of SO2.
You should get the same result using
V = nRT/P , if you use the right R value, P, and T, with 1/64 mole.
Your answer will be in liters, not atm.
THANKS!!!
To calculate the volume of sulfur dioxide gas expelled into the atmosphere per tank of gasoline at STP, you can follow these steps:
1. Determine the number of moles of sulfur and oxygen:
- Convert the mass of sulfur from grams to moles by dividing it by the molar mass of sulfur. The molar mass of sulfur is approximately 32.06 g/mol.
- Convert the mass of oxygen from grams to moles by dividing it by the molar mass of oxygen. The molar mass of oxygen is approximately 16.00 g/mol.
2. Identify the limiting reactant:
- Compare the moles of sulfur and oxygen to determine which reactant is limiting. The reactant that produces a lesser amount of product is the limiting reactant in this case.
3. Use the balanced chemical equation to determine the moles of sulfur dioxide produced:
- The balanced chemical equation for the combustion of sulfur is:
Sulfur + Oxygen -> Sulfur Dioxide
- In this equation, the stoichiometric coefficient for sulfur is 1, and for sulfur dioxide, it is also 1. Therefore, the moles of sulfur dioxide produced will be equal to the moles of sulfur.
4. Calculate the volume of sulfur dioxide gas at STP:
- Use the ideal gas law equation: PV = nRT
P: Pressure (1 atm)
V: Volume (unknown)
n: Number of moles of sulfur dioxide
R: Ideal gas constant (0.0821 L·atm/(mol·K))
T: Temperature (273.15 K)
- Rearrange the equation to solve for V:
V = (nRT) / P
5. Plug in the values and perform the calculation:
- Substitute the number of moles of sulfur dioxide, the ideal gas constant (R), the temperature (T), and the pressure (P) into the equation and calculate the volume.
Note: When using atomic weights, you should consider the atomic weight of one sulfur (S) atom and two oxygen (O) atoms since there are two oxygen atoms present in one molecule of sulfur dioxide (SO2).
The resulting volume will be in liters.