Vanadium metal, V, which is used in steel alloys can be obtained by reacting vanadium(V)oxide, V2O5,with calcium, Ca, at high temperature : 5Ca(s) + V2O5(s) → 5CaO(s) + 2V(s)

Question

When 1.71 x 103 g of an impure sample of calcium reacts with 1.01 x 102 g of vanadium(V)

oxide, 50.0 g of vanadium was yielded. [Ar V: 50.94 g mol-1]

a) Calculate the number of moles of vanadium yielded.

b) Determine the limiting reactant.

c) Determine the percentage of calcium in the impure sample.

Answer 1

5Ca(s) + V₂O₅(s) ==> 5CaO(s) + 2V(s) ... balanced equation

a) moles V produced:

50.0 g V x 1 mol V / 50.94 g = 0.981 mols V

b) To find the limiting reactant, we must first find mols of Ca present and mols of V₂O₅ present.

Since the starting material is 1.71x10³ g of impure Ca, there is no way for us to know the mols of Ca present and so we cannot determine the limiting reactant. If we knew the % Ca in this impure sample, then we could find the mols and hence find the limiting reactant, but how can we find the % Ca without knowing the yield of CaO or the limiting reactant?

mols V₂O₅ used = 101 g V2O5 x 1 mol V₂O₅ / 181.9 g = 0.555 mols V₂O₅

mols Ca used = ?

Can't find limiting reactant without knowing mols Ca used.