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 %