If 3.15 g of n2h4 reacts and produces 0.450 l of n2, at 295 k and 1.00 atm, what is the percent yield of the reaction?

First thing to do is to establish the balanced chemical reaction which is as follows:

3N2H4 = N2 + 4NH3

We need to determine the theoretical yield which means the amount of N2 produced when all of the reactant is used up. We do as follows:

3.15 g N2H4 ( 1 mol / 32.06 g ) ( 1 mol N2 / 3 mol N2H4 ) = 0.0328 mol N2

Assuming ideal gas, 

V = nRT / P = (0.0328)(0.08205)(295) / 1 = 0.7927 L N2

Percent yield = 0.450 / 0.7927 x 100 = 57%

Hope this helps.

The answer is 323 g.

And a) oxygen i believe