Can someone please answer this question

Question

N₂H₄ + O₂ → N₂ + 2H₂O(l)

If 9.55 g of N₂H₄ reacts with excess oxygen and produces 0.400 L of N₂ at 298 K and 1 atm, what is the percentage yield of the reaction?

Answer 1

Stoichiometry steps:

1. Turn quantity of reactant into moles: divide 9.55 g by the molar mass of N2H4 to get .298 moles
2. Figure out the limiting reactant: in this case there is excess oxygen so we don't have to worry about limiting reactant
3. Use the balanced equation to get moles of product: .298 moles N2H4 * 1 mole N2 / 1 mole N2H4 = .298 moles N2
4. Convert moles of product into quantity: this time we need a volume so we use the ideal gas law, PV = nRT. Plug in n=.298 moles, T = 298 K, P = 1 atm and R = .0821 L*atm/K*mol and solve for volume. You should get 7.287 L. This is the predicted yield.
5. Get percent yield: divide your actual yield by the predicted yield, then multiply by 100 to turn the decimal into a percent: .400/7.287 * 100 = 5.49%