N2H4(ℓ) + O2(g) → N2(g) + 2 H2O(g). How many grams of water are produced when 15 g of pure hydrazine is ignited in the presence of 31 g of pure oxygen?

Question

Hydrozine (N2H4) emits a large quantity of energy when it reacts with oxygen, N2H4(ℓ) + O2(g) → N2(g) + 2 H2O(g). How many grams of water are produced when 15 g of pure hydrazine is ignited in the presence of 31 g of pure oxygen?

The answer should be in Grams

Please help urgently, it is due in one-two hours, it would be much appreciated it is my last chance to write the correct answer, I have tried it 4 times but its keep on coming incorrect.

Answer 1

N₂H₄(ℓ) + O₂(g) → N₂(g) + 2 H₂O(g) ... balanced equation

mols N₂H₄ present = 15 g x 1 mol/32 g = 0.469 mols

mols O₂ present = 31 g O₂ x 1 mol/32 g = 0.969 mols

Limiting reactant = N2H4

grams H₂O produced = 0.469 mols N₂H₄ x 2 mols H₂O / 1 mol N₂H₄ x 18 g H₂O/ mol = 16.9 g H₂O = 17 g H₂O (2 sig figs)