What mass of sulfur msut be used to produce 17.1 L of gaseous sulfur dioxide at STP according to the following equation?
S8 (s) + 8O2 (g) -> 8 SO2 (g)
Answer in units of g
one mole of S8 produces 8*22.4 lters
so you want the following number of moles of S8:
(17.1/(8*22.4))
convert that to grams of S8
To calculate the mass of sulfur required to produce a given volume of sulfur dioxide gas, we need to use the stoichiometry of the balanced chemical equation.
First, let's determine the number of moles of sulfur dioxide gas using the ideal gas law at standard temperature and pressure (STP). STP conditions are defined as a temperature of 273.15 K (0°C) and a pressure of 1 atmosphere (atm). The ideal gas law equation is as follows:
PV = nRT
Where:
P = Pressure = 1 atm
V = Volume = 17.1 L
n = Number of moles
R = Ideal gas constant = 0.0821 L·atm/(K·mol)
T = Temperature = 273.15 K
Rearranging the ideal gas law equation to solve for the number of moles:
n = PV / RT
Substituting the given values:
n = (1 atm)(17.1 L) / (0.0821 L·atm/(K·mol))(273.15 K)
n ≈ 0.661 mol (rounded to three decimal places)
According to the balanced chemical equation, we need 8 moles of sulfur dioxide gas (SO2) for every 1 mole of sulfur (S8) reactant.
Therefore, the number of moles of sulfur required can be calculated using the following ratio:
moles of sulfur = (8 moles SO2) × (0.661 mol S / 1 mol SO2)
moles of sulfur = 5.288 mol S (rounded to three decimal places)
Finally, converting moles of sulfur to grams using the molar mass of sulfur (32.07 g/mol):
mass of sulfur = (5.288 mol S) × (32.07 g/mol)
mass of sulfur ≈ 169.626 g (rounded to three decimal places)
Therefore, approximately 169.626 grams of sulfur must be used to produce 17.1 L of gaseous sulfur dioxide at STP.