What is a limiting reagent? In the reaction $N_{2} + 3 H_{2} \to 2 N H_{3}$ , if $20$ g of dinitrogen reacts with $10$ g of dihydrogen what is the mass of ammonia produced. (at. Mass of $N = 14$ ).

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Solution

Limiting Reagent
The reactant which gets completely consumed during the course of reaction is called Limiting Reagent
$N_{2} + 3 H_{2} ⟶ 2 N H_{3}$
$20 g$ $10 g$
No. of mole= $\frac{G i v e n M a s s}{M o l a r M a s s} = n = \frac{x}{M} N_{2}, M o l a r M a s s = 28 g {m o l}^{- 1} n = \frac{x}{M} = \frac{20}{28} = \frac{5}{7} H_{2}, M o l a r M a s s = 2 g {m o l}^{- 1} n = \frac{x}{M} = \frac{10}{2} = 5$
The ratio $\frac{m o l e}{S l i o c h i o m e t r y C o e f f}$ is lowest ie. the limiting reagent
$N_{2} + 3 H_{2} ⟶ 2 N H_{3}$
$\frac{5}{7}$ mol $5$ mol
$N_{2} = \frac{5 / 7}{1} = \frac{n}{S . C .} H_{2} = \frac{5}{3} = \frac{n}{S . C .} \frac{5}{3} > \frac{5}{7}$
Thus $N_{2}$ is the limiting reagent
$1 m o l o f N_{2} ⟶ 2 m o l e s o f N H_{3} \frac{5}{7} m o l o f N_{2} ⟶ \frac{5}{7} \times 2 m o l o f N H_{3} ⟶ \frac{5}{7} m o l o f N H_{3}$
$N H_{3}$
Molar mass= $17 g$
No. of mole= $\frac{10}{7}$
$n = \frac{x}{M} \frac{10}{7} = \frac{x}{17} \frac{10}{7} \times 17 = x \frac{170}{7} = x 24.28 g$
Mass of $N H_{3}$ produced is $24.28 g$

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Similar questions

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?

N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)

6

H_{2}

14

N_{2}

N H_{3}

g

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