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# How many moles of chlorine gas can be produced if 4 moles of FeCl3 react with 4 moles of O2? ?

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Initial moles of FeCl₃ = 4 mol

Initial moles of O₂ = 4 mol

Balanced equation for the reaction:

4FeCl₃ + 3O₂ → 2Fe₂O₃ + 6Cl₂

Mole ratio FeCl₃ : O₂ = 4 : 3

Obviously, when 4 mol of FeCl₃ completely reacts, O₂ needed = 3 mol < 4 mol

Hence, O₂ is in excess and FeCl₃ is the limiting reactant (limiting reagent).

According to the above equation, mole ratio FeCl₃ : Cl₂ = 4 : 6

Moles of Cl₂ gas produced = (4 mol) × (6/4) = 6 mol

• 4 FeCl3 + 3 O2 → 2 Fe2O3 + 6 Cl2

4 moles of FeCl3 would react completely with 3 moles of O2, but there is more O2 present than that, so O2 is in excess and FeCl3 is the limiting reactant.

(4 mol FeCl3) x (6 mol Cl2 / 4 mol FeCl3) = 6 mol Cl2

• 4 FeCl3 (s) + 3 O2 (g) -> 2 Fe2O3 (s) + 6Cl2 (g)

4 moles of O2(4 FeCl3/3 O2)= 5.333 moles of  FeCl3

5.333 moles of FeCl3      x   (6 moles Cl2 (g)/4 moles  FeCl3 (s))

= 8 moles Cl2(g)