Answer:
Fe²⁺ = 80mol/Ls
H⁺ = 50mol/Ls
Mn²⁺ = 4.0x10²mol/Ls
Fe³⁺ = 80mol/Ls
H₂O = 1.0x10²mol/Ls
Step-by-step explanation:
Based on the reaction:
MnO₄⁻ + 5 Fe²⁺ + 8H⁺ → Mn²⁺ + 5 Fe³⁺ + 4H₂O
Where 1 mole of permangante with 5 moles of Fe²⁺ and 8 moles of H⁺ produce 1 mole Mn²⁺, 5 of Fe³⁺ and 4 of H₂O
If the diappearence of 1 mole has a rate of 4.0x10²mol/Ls, disappearence of 5 moles of Fe²⁺ and 8 moles of H⁺ have a rate of:
Fe²⁺ = 1 mole MnO₄⁻ × ( 4.0x10²mol/Ls / 5 moles of Fe²⁺) = 80mol/Ls
H⁺ = 1 mole MnO₄⁻ × ( 4.0x10²mol/Ls / 8 moles of H⁺) = 50mol/Ls
And rate of appearance of 1 mole Mn²⁺, 5 of Fe³⁺ and 4 of H₂O are:
Mn²⁺ = 1 mole MnO₄⁻ × ( 4.0x10²mol/Ls / 1 moles of Mn²⁺) = 4.0x10²mol/Ls
Fe³⁺ = 1 mole MnO₄⁻ × ( 4.0x10²mol/Ls / 5 moles of Fe³⁺) = 80mol/Ls
H₂O = 1 mole MnO₄⁻ × ( 4.0x10²mol/Ls / 4 moles of H₂O) = 1.0x10²mol/Ls