Question

Consider the following oxidation-reduction reaction between vanadium metal and oxygen gas:

4 V(s) + 5 O₂(g)   ?   2 V₂O₅(s)

Calculate the theoretical yield of vanadium(V) oxide (in grams) if you start the reaction with 13.5 grams of vanadium metal (V) as well as 13.5 grams of oxygen gas (O₂). Also, identify the limiting reagent. You must show how you arrived at both of these answers by clearly displaying all calculations below in order to receive credit.

_____ grams of V₂O₅ is the theoretical yield.
_____ is the limiting reagent.

The actual yield was reported as 18.4 g V₂O₅. What was the percent yield?

This experiment was carried out in _____% yield.

Answer 1

Moles of vanadium metal = mass/molecular weight

= 13.5g / 50.94g/mol

= 0.265 mol

Moles of O2 = 13.5/32 = 0.422 mol

From the stoichiometry of the reaction

4 mol of V required = 5 mol O2

0.265 mol of V required = 5*0.265/4 = 0.331 mol O2

But we have more moles of O2 than required 0.331 mol

Excess reactant = O2

Limiting reactant = V

Moles of V2O5 formed = 2 mol V2O5 x 0.265 mol V / 4 mol V

= 0.1325 mol

Theoretical yield = 0.1325 mol x 181.88 g/mol

= 24.1 g

Actual yield = 18.4 g

% yield = actual yield x 100 / Theoretical yield

= 18.4*100/24.1

= 76.35 %