Question

What is the maximum amount of ammonia that can be produced when 0.35 mol Nz reacts with 0.90 mol Hz?

Answer 1

Answer:

0.6 moles NH₃

Explanation:

The reaction that takes place is:

• N₂ + 3H₂ → 2NH₃

First we determine the limiting reactant:

• 0.35 mol N₂ would react completely with (3*0.35) 1.05 moles of H₂. There are not as many H₂ moles, so H₂ is the limiting reactant.

Then we convert H₂ moles (the limiting reactant) to NH₃ moles, keeping in mind the stoichiometry of the reaction:

• 0.90 mol H₂ * $\frac{2molNH_3}{3molH_2}$= 0.6 moles NH₃