JEE Main · 2022 · Shift-ImediumMOLE-203

N2(g) + 3H2(g) <= 2NH3(g); 20 g N2 reacts with 5 g H2. The limiting reagent and moles of NH3 formed are:

Some Basic Concepts (Mole Concept) · Class 11 · JEE Main Previous Year Question

Question

$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$ ; 20 g $\ce{N2}$ reacts with 5 g $\ce{H2}$ . The limiting reagent and moles of $\ce{NH3}$ formed are:

Options

a
$\ce{H2}$ , 1.42 mol
b
$\ce{H2}$ , 0.71 mol
✓
c
$\ce{N2}$ , 1.42 mol
d
$\ce{N2}$ , 0.71 mol

Correct Answerb

$\ce{H2}$ , 0.71 mol

Detailed Solution

🧠 The Haber Process Limit Every $28\,g$ of Nitrogen needs $6\,g$ of Hydrogen ( $1:3$ mole ratio). With $20\,g$ of Nitrogen and $5\,g$ of Hydrogen, you need to check which one runs out first. Nitrogen ( $20\,g$ ) is less than one mole, while Hydrogen ( $5\,g$ ) is more than two moles.

🗺️ Strategic Roadmap Step 1: Evaluate Moles.

N2: $20/28 = 0.714\,mol$ .
H2: $5/2 = 2.5\,mol$ .

Step 2: Find the Limiter. Ratio needed: $0.714\,N2$ : $2.14\,H2$ . We have $2.5\,H2$ . So Nitrogen is the bottleneck.

Step 3: Calculate Ammonia Output. $1$ mole Nitrogen $ightarrow$ $2$ moles Ammonia. $\text{Yield} = 0.714 \times 2 = 1.428\,mol$

$\boxed{\text{Answer: (a) N2, 1.43 mol}}$

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