# How many moles of NH_3 can be produced from 21.0 mol of H_2 and excess N_2?

Jun 30, 2016

Clearly, $14 \cdot m o l$ of ammonia could be produced.

#### Explanation:

The stoichimetric equation:

${N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) \rightarrow 2 N {H}_{3} \left(g\right)$

The stoichiometric equation explicitly states that each mol, each equiv, dinitrogen, requires 3 mol of dihydrogen.

Given that we have $21.0 \cdot m o l \text{ dihydrogen}$, $14 \cdot m o l$ of ammonia could be produced given stoichiometric reaction. Of course, in this reaction, stoichiometric yields are unthinkable, but industrial procedures can recycle the unreacted dinitrogen and dihydrogen and achieve acceptable turnovers. That ammonia is condensable, whereas the reactant gases are much less so, allows recycling.