(at 398 K and 1.20 atm). the molar mass of FeS2 is 119.99 g/mol?
answer 2
Assume the behavior of an ideal gas. STP is 298K and 1.0 atm. First, determine which is the limiting reactant:
PV = nRT
(1 atm)(55.0 L O2) = n * (0.0821 L atm/Kmol)(298 K)
n = 2.25 mol O2 available for the reaction.
(96.7 g FeS2) (1 mol FeS2/119.99 g) = 0.806 mol FeS2 available for reaction.
Using the FeS2 mole figure and stoichiometric equivalences:
(0.806 mol FeS2)(11 mol O2/4 mol FeS2) = 2.22 mol O2
FeS2 is therefore the limiting reagent and 2.22 mol of O2 will react with the available quantity.
For every 11 mol of O2 consumed, 8 mol of SO2 are produced. So,
(2.22 mol O2)(8 mol SO2/11 mol O2) = 1.61 mol SO2
To find the volume of SO2 produced from the amount (moles), again use the ideal gas law:
(1 atm) * V = (1.61 mol)(0.0821 L atm/Kmol)(298 K)
V=39.4L
answer 3
V=32.9L
Answer 1
4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2
(96.7 g FeS2) / (119.9762 mol FeS2/mol) = 0.80599 mol FeS2
(55.0 L O2) x (273 / 398) x (1.20 / 1.00) / (22.414 L/mol) = 2.020 mol O2
0.80599 mol of FeS2 would completely react with 0.80599 x (11/4) = 2.2165 mol of O2, but there isn’t much O2, so O2 is the limiting reactant.
(2.020 mol O2) x (8/11) x (22.414 L SO2/mol) = 32.9 L SO2 in PBS