Question

1)Calculate the number of moles of 2,4-pentanedione (volume is 4.25ml, GMM: 100.11 g mol^-1).

2)Claculate the theoretical yield, in moles of Fe(C5H7O2)3

3.) Calculate the theoretical yield, in grams of Fe(C5H7O2)3

Answer 1

V = 4.25 ml

MW = 100.11 g/mol

Density = 0.976 g/ml

mass = D*V = 4.25 ml * 0.976 g/ml = 4.148 grams of substance

Mol = mass/MW = 4.148/100.11 = 0.04143 mol

2) and 3) are imopssible to solve with more data

BUT, i will assume that the amount of 2,4-pentanedionewas the limiting reactant... therefore:

3 mol of 2,4-pentanedione will produce 1 mol of Fe(C5H7O2)3

We have 0.04143 mol of substance, therefore 1/3 must be prouced, that is, 0.01381 mol of Fe(C5H7O2)3

For that in grams... Mass = mol*MW = 0.01381*100.11 = 1.3825 grams of Fe(C5H7O2)3