You have #28*g# dinitrogen, and #24*g# of dihydrogen. What is the limiting reagent with respect to ammonia formation?

Question

You have #28*g# dinitrogen, and #24*g# of dihydrogen.  What is the limiting reagent with respect to ammonia formation?

Owen Ambrose · Accepted Answer

Dihydrogen is in excess, so dinitrogen is the limiting reagent. Approx. 34 g of ammonia can be produced.

#N_2(g) + 3H_2(g) rarr 2NH_3(g)# You have approx #1# mol #N_2# and #12# mol #H_2#. If you assume 100% yield, 2 mol ammonia should result. Of course, the actual production of ammonia gets nowhere near these yields.

Owen Ambrose · Answer

undefined To determine the limiting reagent, we need to compare the number of moles of each reactant to the stoichiometric coefficients in the balanced equation for the formation of ammonia. The balanced equation is:

N₂ + 3H₂ → 2NH₃

Calculate the number of moles of each reactant:

For dinitrogen (N₂):
$28 g \times \frac{1 \text{ mol}}{28 \text{ g}} = 1 \text{ mol}$

For dihydrogen (H₂):
$24 g \times \frac{1 \text{ mol}}{2 \text{ g}} = 12 \text{ mol}$

Since the stoichiometric ratio between N₂ and H₂ is 1:3, we need 3 moles of H₂ for every 1 mole of N₂. However, we have only 1 mole of N₂ available, which means H₂ is in excess. Therefore, the limiting reagent is dinitrogen (N₂).