3. If 25.5 g H2 are made to react with 64.2 g N2, what is the limiting reagent? 3H2(g) + N2(g) → 2NH3(g)

Question

3. If 25.5 g H2 are made to react with 64.2 g N2, what is the limiting reagent? 3H2(g) + N2(g) → 2NH3(g)

Answer 1

First, write a correctly balanced equation:

3H₂(g) + N₂(g) ==> 2NH₃(g)

Next, to find the limiting reagent, one way is to find the moles of each reactant and divide them by the corresponding coefficient in the balanced equation. Whichever value is less will represent the limiting reactant.

For H₂ we have: 25.5 g H₂ x 1 mole / 2 g = 12.8 moles (÷3 ->4.3)

For N₂ we have: 64.3 g N₂ x 1 mole / 28 g = 2.30 (÷2 ->1.15)

Since 1.15 is less than 4.3, we now know that N₂ is the limiting reactant

Answer 2

Basically the limiting reagent is the one that runs out faster.

We first write the balanced equation and then calculate the number of moles we have for each reactant.

Balanced eq: 3H2​(g)+N2​(g)→2NH3​(g)

mole H2: 25.5 g/ 2g/mol = about 12 moles

Mole N2: 64.2 g/ 28 g/mol = about 2.3 moles

Going based on our balanced equation, for every mole of N2, we need 3 moles of H2. So we need 2.3 moles of N2 times 3 = about 7 moles of H2. We need 7 moles of H2 to finish all our N2 but have 12 moles of H2 instead (some leftover for H2). We can say N2 is our limiting reagent then because it is used up first in reaction with nothing left. (Stop reaction from continuing)