What is the theoretical yield of vanadium that can be produced by the reaction of 40.0 g of V₂O₅ with 40.0 g of calcium based on the following chemical reaction
V₂O₅(s) + 5Ca(l) $\rarr$ 2V(l) + 5CaO(s)

Question

What is the theoretical yield of vanadium that can be produced by the reaction of 40.0 g of V₂O₅ with 40.0 g of calcium based on the following chemical reaction
V₂O₅(s) + 5Ca(l)

\rarr

2V(l) + 5CaO(s)

Quizplus · Accepted Answer

To calculate the theoretical yield, we need to first determine the limiting reactant. We can do this by converting the mass of each reactant into moles. 40.0 g V₂O₅ × (1 mol/259.30 g) = 0.154 mol V₂O₅ 40.0 g Ca × (1 mol/40.08 g) = 0.999 mol Ca According to the balanced equation, we need 5 moles of Ca for every 1 mole of V₂O₅. This means that Ca is in excess and V₂O₅ is the limiting reactant. We can now use the mole ratio from the balanced equation to calculate the theoretical yield of V. 0.154 mol V₂O₅ × (2 mol V/1 mol V₂O₅) × (50.94 g V/1 mol V) = 20.3 g V Therefore, the theoretical yield of vanadium that can be produced is 20.3 g. The best choice is C.

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