If 2.70 L of hydrogen gas reacts completely with excess nitrogen gas by this reaction, how many grams of ammonia are produced? The reaction is carried out at 3.00 atm and 298 K.

Question

Ammonia can be produced from hydrogen and nitrogen gases based on the reaction

N2(g) + 3H2(g) à 2NH3(g)

If 2.70 L of hydrogen gas reacts completely with excess nitrogen gas by this reaction, how many grams of ammonia are produced? The reaction is carried out at 3.00 atm and 298 K.

Answer 1

N₂(g) + 3H₂(g) ==> 2NH₃(g) ... BALANCED EQUATION

Use the Ideal gas law to find the moles of H₂(g):

PV = nRT

n = PV/RT = (3.00 atm)(2.70 L) / (0.0821 Latm/Kmol)(298K)

n = 0.331 moles H2 gas

Use the stoichiometry of the balanced equation to find moles of NH₃(g) fomred:

0.331 moles H₂ x 2 moles NH₃ / 3 mols H₂ = 0.221 moles NH₃ formed

Use molar mass of NH₃ (17 g / mol) to convert this to grams of NH₃ formed:

0.221 mols NH₃ x 17 g / mol = 3.76 g NH₃ formed