Question

200 gm N2 is mixed with H2 and gives _____.how much gram. NH3 and which limiting reagent in this reaction will be?​

Answer 1

Answer:

Balanced Chemical Equation:

⇒ N₂ + 3H₂ = 2NH₃

1 mole of Nitrogen reacts with 3 moles of Hydrogen to form 2 moles of Ammonia.

Given Mass of Nitrogen = 200 g

Molecular Mass of Nitrogen = 28 g

Therefore number of moles (n) = 200g / 28g

⇒ n(N) = 7.14 moles of Nitrogen

Given mass of Hydrogen = 200 g

Molecular Mass of Hydrogen = 2 g

Therefore no of moles (n) = 200g / 2g

⇒ n(H) = 100 moles of Hydrogen

According to the balanced chemical reaction, 1 mole of Nitrogen reacts with 3 moles of Hydrogen. Hence 7.14 moles of Hydrogen would react with:

$\implies x = \dfrac{7.14 \times 3}{1} = \boxed{\bf{21.42\textbf{ moles of Hydrogen}}}$

But since we have 200 moles of Hydrogen which is in excess, the limiting reagent is Nitrogen.

(Limiting Reagent refers to the substance which is present in lesser quantity which gets depleted completely once the reaction is over.)

Hence if 1 mole of Nitrogen gives 2 moles of Ammonia, 7.14 moles of Nitrogen will give:

$\implies y = \dfrac{2 \times 7.14}{1} = \boxed{\bf{14.28\textbf{ moles of Ammonia}}}$

In terms of grams, we get:

⇒ 14.28 × Molar Mass of Ammonia

⇒ 14.28 × 17 = 242.76 grams of Ammonia