What is the limiting reactant

Question

What reactant is limiting if 4.52 g of ammonium nitrate and 3.87 grams of sodium hydroxide are used?

Answer 1

First write the correctly balanced equation for the reaction taking place:

NH₄NO₃(aq) + NaOH(aq) ==> NH₄OH(aq) + NaNO₃(aq) ==> NH₃(g) + H₂O(l) + NaNO₃(aq)

So, the overall summary reactions would be....

NH₄NO₃(aq) + NaOH(aq) ==> NH₃₍g) + NaNO₃(aq)

To find the limiting reactant, we find the moles of NH4NO3 present and the moles of NaOH present and sinc they react in a 1:1 ratio as seen from the balanced equation, whichever is present in the lowest number of moles will be limiting.

moles NH₄NO₃ = 4.52 g x 1 mole/80.04 g = 0.0565 moles

moles NaOH = 3.87 g x 1 mole/40.0 g = 0.0968 moles

The NH₄NO₃ is limiting