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Sagot :
The reaction between 96.7 g of fes2 and 55.0 l of o2 produced the volume of so2 (at stp) (at 398 k and 1.20 atm). Fes2's molar mass is 119.99 g/mol, which is equivalent to 32.94 L.
FeS2 weighs 96.7 g.
0.806 mol of FeS2 is equal to 96.7 g of mass divided by 119.99 g of molar mass.
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Given 55.0 L of oxygen. We need to find the volume of Oxygen at STP as we need to determine the volume of SO2 at STP.
Hence, we use the equation, P1V1/T1 = P2V2/T2
(1.20 x 55.0) / 398 = (1.0 x V2) / 273
V2 = 45.27 L
Thus, volume of oxygen = 45.27 L
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Now, we need to find moles of oxygen using the equation PV = nRT
(1.0 x 45.27) = n (0.0821 x 273)
n = 2.02 mol
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Thus, the reactants we have are
Moles of FeS2 = 0.806 mol
Moles of oxygen = 2.02 mol
Now, we need to find the limiting reagent. In order to find the limiting reagent we need to divide the moles we found with the respective stoichiometric coefficients.
Thus, moles of FeS2 = 0.806 mol / 4 = 0.2015 mol
Moles of O2 = 2.02 mol / 11 = 0.184 mol.
As, the moles of O2 are less than that of FeS2, O2 is the limiting reagent.
Hence, we use moles of O2(2.02 mol) to find the SO2 produced.
From the chemical equation, it is clear that 11 mol of O2 produces 8 mol of SO2.
Thus, 2.02 mol of O2 produces = (2.02 mol O2) x (8 mol SO2 / 11 mol O2) = 1.47 mol of SO2
Using PV = nRT, let us find volume of SO2.
(1.0)(V) = 1.47 x 0.0821 x 273
V = 32.94 L of SO2 is produced.
learn more about fes2 here:
https://brainly.com/question/15583224
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