# If 28 g of N_2 and 25 g of H_2 are reacted together, which one would be the limiting reactant?

May 22, 2017

${N}_{2}$

#### Explanation:

If we assume here the product is ammonia, $N {H}_{3}$, the equation looks like

${N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s 2 N {H}_{3} \left(g\right)$

To find the limiting reactant, let's convert the given masses of ${N}_{2}$ and ${H}_{2}$ to moles using their molar masses, and then divide the calculated number by the coefficient in front of it in the equation. Whichever number in the end is lower, that reactant is limiting.

Using the molar masses of ${N}_{2}$ and ${H}_{2}$:

28cancel(gN_2)((1molN_2)/(28.02cancel(gN_2))) = color(red)(1.0 mol N_2

25cancel(gH_2)((1molH_2)/(2.02cancel(gH_2))) = (12molH_2)/(3"(coefficient)") = color(blue)(4.0 mol H_2

Since ${N}_{2}$ is present in a lower relative amount than ${H}_{2}$, nitrogen is the limiting reactant.