b) Mass of MnO₂ = 5.981 g
Mass of H₂O = 1.2384 g
c) Total Mass of Reactant consumed = 11.708 g
b) Given Reaction
Zn(s) + MnO₂(s) + H₂O(l) → ZnO(s) + Mn(OH)₂(s)
Mass of Zn = 4.50 g
Moles of Zn = 0.0688 moles
Now,
Moles of Zn = moles of MnO₂ = moles of H₂O = moles of ZnO = moles of Mn(OH)₂
Hence ,
Moles of MnO₂ = 0.0688 moles
Mass of MnO₂ = 0.0688 × 86.9368 g
= 5.981 g
Similarly,
Moles of H₂O = 0.0688 moles
Mass of H₂O = 0.0688 × 18 g
= 1.2384 g
c) now ,
Moles of ZnO = 0.0688 moles
Mass of ZnO = 0.06880× 81.3794 g
= 5.598 g
Moles of Mn(OH)₂ = 0.0688 moles
Mass of Mn(OH)₂ =0.0688 × 88.952 g
= 6.11 g
Total mass of Product = 11.708 g
Total Mass of Reactant = 11.715 g
Hence,
Total mass of reactant consumed = 11.708 g
c) As total mass of reactant is more than that of mass of reactant consumed , Hence G is more than that of mass of reactant consumed .
G = - nFEcell
and no. of moles of reactant is greater than that of number of moles of reactant consumed .
Hence voltaic cell of given Capacity are heavier than that of mass of reactant consumed .
Thus from above conclusion we can say that , Mass of the reactant consumed is 11.708 g.
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= 80 mmol K⁺