What is the theoretical yield of vanadium, in moles, that can be produced by the reaction of 2.0 mole of V₂O₅ with 6.0 mole of calcium based on the following chemical reaction? V₂O₅(s)+ 5Ca(l)$\rarr$ 2V(l)+ 5CaO(s) A)1.0 mol B)1.6 mol C)2.0 mol D)2.4 mol E)4.0 mol

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From the balanced equation, we can see that 1 mole of V₂O₅ reacts with 5 moles of Ca to produce 2 moles of V. Therefore, if 2.0 moles of V₂O₅ reacts with 6.0 moles of Ca, which is 3 times the amount needed, the limiting reactant will be V₂O₅. Theoretical yield of V can be calculated by using stoichiometry as follows: 2.0 mol V₂O₅ x 2 mol V / 1 mol V₂O₅ = 4.0 mol V Therefore, the theoretical yield of vanadium that can be produced is 4.0 mol, which is option D.

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