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
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To find out how many grams of sulfur dioxide will be produced, we need to use stoichiometry and the ideal gas law.
1. Start by determining the number of moles of O2 using the ideal gas law equation:
PV = nRT
In this case:
P = 1.3 atm
V = 24.0 L
T = 15°C + 273.15 = 288.15 K (temperature converted to Kelvin)
Rearrange the equation to solve for moles (n):
n = PV / RT
Substitute the values:
n = (1.3 atm * 24.0 L) / (0.0821 L·atm/mol·K * 288.15 K)
2. Calculate the moles of SO2 using stoichiometry.
According to the balanced equation, 1 mole of CS2 reacts with 3 moles of O2 to produce 2 moles of SO2.
Since we have the number of moles of O2 from the previous step, multiply it by the stoichiometric ratio:
moles of SO2 = n (O2) * (2 moles SO2 / 3 moles O2)
3. Convert moles of SO2 to grams.
Calculate the molar mass of SO2 (sulfur dioxide):
Molar mass = (1 atom of S * atomic mass of S) + (2 atoms of O * atomic mass of O)
Multiply the moles of SO2 by the molar mass of SO2:
grams of SO2 = moles of SO2 * molar mass of SO2
By following these steps, you should be able to calculate the number of grams of sulfur dioxide produced.