(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?

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Parveen,

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} $

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