Consider the following oxidation-reduction reaction between vanadium metal and o
ID: 704039 • Letter: C
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.
Explanation / Answer
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 %