# Theoretical Yield and Percentage Yield

#### Smokey

2.5 g of V2O5 was dissolved in 10mL of water, 5mL of H2SO4 and 15mL of ethanol. After being heated on a steam bath the solution was filtered and the filtrate was transferred to a 400mL beaker. 6.2 mL of acetylacetone was added and then the solution was neutralised by 10mL of Na2CO3. The product was VO(acac)2. Calculate the theoretical yield and percentage yield if the yield obtained was 2.58 grams.

Equation

V2O5 + 2 H2SO4 + C2H5OH → 2 VOSO4 + 3 H2O + CH3CHO
VOSO4 + 2 C5H8O2 + Na2CO3 → VO(C5H7O2)2 + Na2SO4 + H2O + CO2

Calculations:

V2O5: (2.5 g)/(181.88 g/mol) = 0.013745 mol
C2H5OH: (15 mL x 0.789 g/mL)/(46.06844 g/mol) = 0.2569 mol
H2SO4: (5 mL x 1.84 g/mL)/(98.079 g/mol) = 0.0938 mol /2 = 0.0469

From the number of moles of each reactant and the mole ratios in the balanced equation, you see that V2O5 is clearly the limiting reagent in the first reaction. And since the mole ratio of VOSO4 to V2O5 is 2:1, you would make 2x the moles of VOSO4, or 0.02749 mol. So in the 2nd step the moles of each reactant are:

VOSO4: 0.02749 mol
C5H8O2: (6.2 mL x 0.98 g/mL)/(100.13 g/mol) = 0.06068 mol
Na2CO3: 10 g/(105.9885 g/mol) = 0.0943 mol
You see that VOSO4 is the limiting reagent; since the product and the VOSO4 are in a 1:1 mole ratio you would also make 0.02749 moles of VO(C5H7O2)2.

Theoretical yield = 0.02749 mol x 265.157 g/mol = 7.29 g

Can somebody please double check this for me???? Thanks