Question

If 3.05 g of N2H4 reacts and produces 0.950 L of N2, at 295 K and 1.00 atm, what is the percent yield of the reaction?

Answer 1

The reaction will be as follows--

N₂H₄ + O₂ --------> N₂ + 2H₂O

Mole ratio of N₂:N₂H₄ = 1:1

Now,

(3.05 g N₂H₄) / (32.0452 g N₂H₄/mol) x (1 mol N₂ / 1 mol N₂H₄) = 0.095178 mol N₂ in theoretical

Given that, T = 295 K , P = 1.00 atm

s0, PV= nRT
=> V = nRT / P

=> V = (0.095178 mol) x (0.08206 L atm/K mol) x (295 K) / (1.00 atm)

=> V = 2.3040 L N₂ in theoretical

Also, given the actual yield of N₂ = 0.950 L

Percent yield = (actual yield / theoretical yield) x 100%

=> Percent yield = (0.950 L) / (2.3040 L) x 100 %

=> Percent yield = 0.4123 x 100 % = 41.23 %