Question

One step in the production of vanadium uses the reaction of V20, with an excess of aluminum at high temperature to give vanadium and Al2O3. In a test run, 38.4 kg of vanadium is obtained from 72.0 kg of V205. Calculate the percentage yield of vanadium. Hint: What is the theoretical yield of vanadium?

Answer 1

V2O5 + Al ==> V + Al2O3

Balanced equation is

3 V2O5 + 10 Al ==> 6 V + 5 Al2O3

Given Access Al so V2O5 is limiting reactant

From the reaction we can say that

3 moles of V2O5 produces 6 moles of Vanadium

But given 72 Kg of V2O5

Moles of V2O5 = mass of V2O5/molar mass of V2O5

= 72 * 1000 g/181.88 g/mol

= 395.9 moles

So

395.9 moles V2O5 will produce ( 395.9 moles V2O5*6 moles V/3 moles V2O5)

= 791.8 moles Vanadium is produced

So mass of Vanadium produced = moles of Vanadium * molar mass of Vanadium

= 791.8 moles * 50.9 g/mol

= 40.3 Kg

This is the theoretical yield

So

Percent yield = ( actual yield / theoretical yield) * 100

Given 38.4 Kg Vanadium is obtained

So

Percent yield = (38.4 kg / 40.3 kg) * 100

= 95.2 % is the percentage yield of Vanadium