if 24.0 L of O2 burns completely a sample of carbon disulfide, how many grams of sulfur dioxide will be produced at 15 degrees celcius at a pressure of 1.3 atm?
CS2 + 3O2 -> CO2 + 2 SO2
See other post.
To solve this problem, we need to use the ideal gas law and stoichiometry. The first step is to find the number of moles of oxygen gas (O2) that react.
We know that the volume of the oxygen gas is 24.0 L, and we also need to consider the given temperature and pressure. However, the ideal gas law equation (PV = nRT) requires the temperature to be in Kelvin, so we need to convert the temperature from Celsius to Kelvin by adding 273.15:
T = 15°C + 273.15 = 288.15 K.
Next, we can calculate the number of moles of O2 using the ideal gas law equation:
PV = nRT
(1.3 atm)(24.0 L) = n(0.0821 L·atm/mol·K)(288.15 K)
31.2 atm·L = n(23.62 L·atm/mol)
n = 31.2 atm·L / 23.62 L·atm/mol ≈ 1.32 mol O2.
Now that we know the number of moles of O2, we can use the stoichiometry of the balanced chemical equation to find the number of moles of sulfur dioxide (SO2) produced.
According to the balanced chemical equation: 1 mole of CS2 reacts with 3 moles of O2 to form 2 moles of SO2.
So, since we have 1.32 moles of O2, we can calculate the moles of SO2 as follows:
(1.32 mol O2) × (2 mol SO2 / 3 mol O2) ≈ 0.88 mol SO2.
Finally, we can convert the moles of sulfur dioxide to grams using the molar mass of SO2. The molar mass of sulfur dioxide is approximately 64.07 g/mol.
(0.88 mol SO2) × (64.07 g SO2 / 1 mol SO2) ≈ 56.33 g SO2.
Therefore, approximately 56.33 grams of sulfur dioxide will be produced.