Question

Metallic vanadium, V, can be obtained from its oxide from the reaction below.

V₂O₅(s) + 5Ca(l) = 2V (s) + 5CaO(s)

If 100. g of V₂O₅ is reacted with 500.0 g of Ca, how many grams of V can be produced?

		127 g
		56.0 g
		100. g
		50.9 g

Answer 1

Solution :-

Given data

Balanced reaction equation

V₂O₅(s) + 5Ca(l) ----> 2V (s) + 5CaO(s)

Mass of V₂O₅ = 100 g

Mass of Ca = 500 g

Mass of V = ?

Lets calculate moles of each reactant and find limiting reagent

Formula to calculate moles is as follows

Moles = mass in gram / molar mass

Moles of V₂O₅ = 100.0 g/ 181.879 g per mol

                         = 0.55 mol V₂O₅

Moles of Ca = 500.0 g /40.078 g per mol

                     = 12.5 mol Ca

Using the mole ratio of the reactant find the limiting reactant

Mole rati of the V₂O₅ to Ca is 1:5

So lets calculate moles of the Ca needed to react with 0.55 mol V₂O₅

(0.55 mol V₂O₅* 5 mol Ca) / 1 mol V₂O₅ = 2.75 mol Ca

Since moles of Ca required are less than moles of Ca present so Ca is the excess reagent and V₂O₅ is the limiting reactant.

Now lets calculate moles of the V produced from the limiting reactant moles . using the mole ratio of the V and V₂O₅

(0.55 mol V₂O₅* 2 mol V) / 1 mol V₂O₅ = 1.1 mol V

Now lets convert moles of V to its mass

Mass = moles * molar mass

Mass of V = 1.1 mol V * 50.942 g per mol

                  = 56.0 g V

Therefore mass of the v produced = 56.0 g