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