Question

guyz plz do this ...... question of chemistry........fast with full explanation......

Answer 1

Hey friend, Harish here.

Here is your answer:

Given that:

Amount of N₂ = 50 Kg.

Amount of H₂ = 10 Kg.

To find,

The limiting reagent and the amount of NH₃ formed in the reaction.

Solution:

The reaction is :   N₂ + 3H₂  →  2NH₃ 

Molecular Mass of Nitrogen = 28 g/mol = 0.028 kg/mol

Molecular Mass of Hydrogen = (3 × 2) g/mol = 0.006 kg/mol

Molecular Mass of Ammonia = (2 × 17)g/mol = 0.034 kg/mol

From the reaction we can see that one mole of Nitrogen reacts with 3 moles of Hydrogen to give 2 moles of ammonia.

⇒ 0.028 kg of N₂ reacts with 0.006 kg of H₂ .

⇒ Then,  50 kg of N₂ reacts with x grams of H₂.

By unitary method we can find the value of x.

⇒ $x = \frac{(0.006)\times (50)}{(0.028)} = 10.71\ Kg$

So, We need 10.71 Kg of hydrogen to react with 50 kg of nitrogen .

But we have only 10 kg of H₂. 

Therefore H₂ is the Limiting reagent. 

Then,

0.006 Kg of Hydrogen will produce 0.034 kg of ammonia.

And 10 kg of Hydrogen will produce x kg of ammonia.

So again by unitary method;

⇒  $x = \frac{(0.034)\times (10)}{(0.006)} = 56.667 Kg$

We know that,

$No.\ of\ moles = \frac{Given \ Mass}{Molecular\ mass}$

Here mass must be in grams

⇒ $\frac{56.67 \times 1000}{17} = 3.3 \times 10^{3}$

Therefore , The limiting reagent is H₂. And amount of ammonia formed is 3.3 × 10³
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Hope my answer is helpful to you.