Answer to Question #256659 in General Chemistry for Jonalyn Galao Gamu

Question

V(s) + O2g - V2O3(s) 50.94g/mol +32.00g/mol^ 149.88g/mol A.)calculate the theoretical yield of vanadium (lll) oxide, assuming you begin with 200.00 grams vanadium metal. B.) after the experiment is performed, an experimental yield of 183.2 grams is produced.calculate the percent yield for this experiment. C.) determine the percent error. D.) compute for the excess amount of the excess reactant.

Answer 1

The balanced equation:

4V(s) + 3O₂(g) --> 2V₂O₃(s)

A)

$Theoretical\ yield\ (V_2O_3)=200.00\ g(V)\times\frac{1\ mol(V)}{50.94\ g(V)}\times\frac{2\ mol(V_2O_3)}{4\ mol(V)}\times\frac{149.88\ g(V_2O_3)}{1\ mol(V_2O_3)}=294.2\ g$

Answer: 294.2 g

B)

$\%\ yield=\frac{actual\ yield}{theoretical\ yield}\times100\%=\frac{183.2\ g}{294.2\ g}\times100\%=62.27\%$

Answer: 62.27 %

C)

$\%\ error=\frac{|actual\ yield-theoretical\ yield|}{|theoretical\ yield|}\times100\%=\frac{|183.2\ g-294.2\ g|}{294.2\ g}\times100\%=37.73\%$

Answer: 37.73 %

D)

The excess reactant leftover can only be calculated in case of assumption that the actual yield of the experiment is less than the theoretical yield only due to the shortage of one of the reactants (in this case - O₂ because the amount of V is given), but for no other reason. Therefore, the amount of vanadium metal involved in the reaction can be found:

$m(V)=183.2\ g(V_2O_3)\times\frac{1\ mol(V_2O_3)}{149.88\ g(V_2O_3)}\times\frac{4\ mol(V)}{2\ mol(V_2O_3)}\times\frac{50.94\ g(V)}{1\ mol(V)}=124.5\ g$

So the mass of the excess reactant equals:

$m(V,excess)=200.00\ g-124.5\ g=75.5\ g$

Answer: 75.5 g