Ammonia NH3 gas and oxygen O2 gas react to form nitrogen N2 gas and water H2O vapor. Suppose you have 2.0 mol of NH3 and 13.0 mol of O2 in a reactor. Calculate the largest amount of N2 that could be produced. Round your answer to the nearest 0.1 mol.

Respuesta :

Answer:

1.0 mol N₂

Explanation:

The reaction that takes place is:

  • 4NH₃ + 3O₂ → 2N₂ + 6H₂O

First we determine the limiting reactant:

13.0 mol of O₂ would react completely with (13*4/3) 17.3 moles of NH₃. There are not as many NH₃ moles so NH₃ is the limiting reactant.

Then we calculate the number of N₂ that could be formed from the given moles of NH₃:

2.0 mol NH₃ * [tex]\frac{2molN_2}{4molNH_3}[/tex] =  1.0 mol N₂