$\begin{array}{l}{N}_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)\\ 20g5g\end{array}$ Consider the above reaction, the limiting reagent of the reaction and number of moles of NH₃ formed respectively are:

Limiting Reagent in N₂ + 3H₂ → 2NH₃ Reaction

In this chemical reaction:

N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g

We are given:

• 20 g of N₂
• 5 g of H₂

Step 1: Calculate the Moles of Each Reactant

Moles of N₂ = Mass / Molar Mass = 20 g / 28 g/mol = 0.714 mol

Moles of H₂ = Mass / Molar Mass = 5 g / 2 g/mol = 2.5 mol

Step 2: Identify the Limiting Reagent

From the balanced equation:

N₂ + 3H₂ → 2NH₃

The mole ratio between N₂ and H₂ is 1:3. To react completely with 0.714 mol of N₂, we need:

Required H₂ = 0.714 mol × 3 = 2.142 mol

Since we have 2.5 mol of H₂, which is more than the required amount, H₂ is in excess.

Thus, the limiting reagent is N₂.

Step 3: Calculate the Moles of NH₃ Formed

From the balanced equation:

1 mol of N₂ produces 2 mol of NH₃

Moles of NH₃ formed = 0.714 mol N₂ × 2 = 1.428 mol

Final Answer:

The limiting reagent in the reaction is N₂ and the number of moles of NH₃ formed is 1.428 mol.

By understanding the steps above, you can easily determine the limiting reagent and calculate the number of moles of ammonia formed in the reaction. This is crucial in stoichiometry problems and helps predict product quantities accurately.