# (solved)Question 1.24 of NCERT Class XI Chemistry Chapter 1

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N2 (g) + H2 (g) → 2NH3 (g)(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 ×103 g of dihydrogen. (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass?
(Rev. 18-Jun-2024)

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### Question 1.24 NCERT Class XI Chemistry

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N2 (g) + H2 (g) → 2NH3 (g)(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 ×103 g of dihydrogen. (ii) Will any of the two reactants remain unreacted? (iii) If yes, which one and what would be its mass?

### Video Explanation(detailed solution given after this video)

Please watch this youtube video for a quick explanation of the solution:

### Solution in Detail

Start by writing the molar masses of the reactants and products.

Molar mass of N2 = 14.00 x 2 = 28.00 g/mol

Molar mass of H2 = 1.008 x 2 ≈ 2.00 g/mol

Molar mass of NH3 = 14.00 + 1.008 x 3 ≈ 17.00 g/mol

\begin{aligned} &\underline{\text{Step 1: What is Given}}\\ &\text{N}_2 = 2.00 \times 10^3 \text{ g}\\ &\text{H}_2 = 1.00 \times 10^3 \text{ g}\\\\ &\underline{\text{Step 2: Convert to Mols}}\\\\ &\text{n} = \frac{\text{mass}}{\text{Molar Mass}}\\\\ &\text{mols of N}_2 = \frac{2.00 \times 10^3}{28.00} = \frac{500}{7} \text{ mol}\\\\ &\text{mols of H}_2 = \frac{1.00 \times 10^3}{2.00} = 500 \text{ mol}\\\\ &\underline{\text{Step 3: Find the Limiting Reagant}}\\ &\text{the balanced reaction is}\\ &\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3\\\\ &\implies 1 \text{ mol N}_2 \text{ reacts with }3\text{ mol H}_2\\\\ &\therefore \frac{500}{7} \text{ mol N}_2 \equiv 3 \times \frac{500}{7} = 214 \text{ mol H}_2 \\\\ &\text{but we have } 500 \text{ mol H}_2 \text{ i.e., excess}\\ &\text{hence N}_2 \text{ is consumed and is the limiting reagant}\\\\ &\underline{\text{Step 4: Mass of Ammonia Produced}}\\\\ &\text{Molar mass of NH}_3 = 17 \text{ g/mol}\\\\ &\text{N}_2 \text{ is limiting reagant given } = 2000 \text{ g}\\ &\therefore \text{ we calculate on basis of N}_2\\\\ &\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3\\\\ &\implies 1 \text{ mol N}_2 \equiv 2 \text{ mol NH}_3\\\\ &\implies 28 \text{ g N}_2 \equiv (2 \times 17) = 34\text{ g NH}_3\\\\ &\implies 2000 \text{ g N}_2 \equiv \bigg(\frac{34}{28}\bigg) \times 2000 \text{ g NH}_3\\\\ &=2.43 \text{ kg ammonia } \:\underline{Ans} \end{aligned}