Calcule la masa en gramos de FeS formada cuando 9.42 g de hierro reacciona con 8.50 g de azufre de acuerdo con la siguiente reacción.

Fe(s) + S(s) > FeS(s)​

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.

Moles of co₂ = 2.35/44 = 0.0534 mole.from ideal gas equation pv=nrt,  t = 18+273k = 291 kelvin.∴px5 = 0.0534x0.0821x291∴p = 0.0534x0.0821x291/5 = 0.2551 atm.now, 1 atm = 760 mm of hg  ∴ 0.02551 atm = 0.02551 x 760 = 19.38 mm of hg

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