How many grams of ammonia (NH3) would be produced when 35.0 grams of nitrogen react with excess hydrogen?

Answer 1

N₂(g) + 3H₂(g) ==> 2NH₃(g) ... balanced equation

35.0 g N₂ x 1 mol N₂ / 28 g x 2 mol NH₃ / mol N₂ x 17 g NH₃ / mol NH₃ = 42.5 g NH₃

Answer 2

First balance the equation

N₂+2H₂ --> 2NH₃

Now determine the molar mass of Nitrogen and Ammonia

Now, do stoichiometry

(35gN₂/1)(1mol N₂ / 28g N₂) = 1.25mol N₂ (2mol NH₃ / 1mol N₂) = 2.5 mol NH₃ (17g NH₃ / 1mol NH₃) = 42.5g NH₃