If 3.95 g of n2h4 reacts and produces 0.750 l of n2, at 295 k and 1.00 atm, what is the percent yield of the reaction?

Molar mass of N2H4 = 32 grams/mole 
3.95 grams of  N2H4 = 3.95/32
                                  = 0.123 moles 

This will produce 0.123 moles of N2 
Now,
From the gas law equation. 
P.V = n x R x T 
P = 1 atm (given)
V = 0.123 x 0.082057 x 295 
V = 2.97 Liters 

Theoretical yield = 2.97 Liters.
Actual yield = 0.750 Liters 

percentage yield = (0.75/2.97) x 100 %
                           = 25.25 %