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A 22.4 L high pressure reaction vessel is charged with 0.5020 mol of iron powder and 1.39 atm of oxygen gas at standard temperature. On heating, the iron and oxygen react according to the balanced reaction below. 4Fe(s) + 3O2(g) → 2Fe2O3(s) After the reaction vessel is cooled, and assuming the reaction goes to completion, what pressure of oxygen remains?

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Answer:

1.013 atm of oxygen remain.

Explanation:

4Fe(s) + 3O₂(g) → 2Fe₂O₃(s)

If the problem asks us to calculate the pressure of oxygen remaining, it would mean that O₂ is the excess reactant, to confirm that:

  • We calculate the moles of O₂ using PV=nRT, knowing that standard temperature is 273.15 K
  • 1.39 atm * 22.4 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.15 K
  • n = 1.39 mol O₂
  • With 1.39 mol O₂ we would need 1.39*4/3 = 1.85 mol Fe to completely deplete O₂, there's not enough iron, so O₂ is indeed the excess reactant.

The moles remaining of O₂ are equal to the original moles minus the reacting moles, which are calculated using the available moles of Fe:

  • 0.5020 mol Fe * [tex]\frac{3molO_{2}}{4molFe}[/tex] = 0.3765 mol O₂

Thus the remaining moles of O₂ are:

  • 1.39 - 0.3765 = 1.0135 mol O₂

Finally we use PV=nRT again to calculate the new value of P:

  • P * 22.4 L = 1.0135 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.15 K
  • P = 1.013 atm

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