Question

Answer the following Questions. [Each carries 4 marks
50.0 kg of N2(g) and 10.0 kg of H2(g) are mixed to produce NH3(g). Calculate the NH3(g) formed.
Identify the limiting reagent in the production of NH, in this situation.​

Answer 1

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✤ Required Answer:

✒ GiveN:

• Amount of N2 = 50 kg
• Amount of H2 = 10 kg
• They are mixed to produce NH3

✒ To FinD:

• Amount of NH3 formed....?
• Limiting reagent in this reaction....?

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✤ How to solve?

First of all, Let's know about the Limiting reagent and excess reagent.

• Limiting reagent - The reagent which are used up in a reaction is known as Limiting reagent.
• Excess reagent - And the reagent that remains unused (Some amount) is known as Excessive reagent.

➤ Formula for finding mole:

$\large{ \boxed{ \sf{No. \: of \: moles = \frac{given \: weight}{molecular \: weight} }}}$

☕By using these concepts, let's solve....After solving, You wills automatically get the explanation of above concept.

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✤ Solution:

Firstly, Let's find moles of reactants:

We have,

• Weight of N2 = 50 kg
• Molecular weight = 28 g

By using formula,

➙ Moles of N2 = 50 × 10³ / 28

➙ Moles of N2 = 17. 86 × 10² moles.

Then,

We have,

• Weight of H2 = 10 kg
• Molecular weight of H2 = 2 g

By using formula,

➙ Moles of H2 = 10 × 10³ / 2

➙ Moles of H2 = 5 × 10³ moles.

☯️ Reaction:

$\large{ \sf{N_2 + 3H_2 \longrightarrow \: 2NH_3}}$

If 1 mole of N2 and 3 moles of H2 reacts then 2 mole of NH3 is produced.

➙ 1 mole of N2 = 3 moles of H2

➙ 17.86 × 10² moles of N2 = 17.86 × 10² × 3 moles of H2

➙ 17.86 × 10² moles of N2 = 5.36 × 10³ moles of H2

But,

• H2 ➝ 5 × 10³ moles present

☀️ So, H2 is the limiting reagent because all the H2 will get utilised in the reaction. N2 is the excess reagent.

So, we know,

➙ 3 mole of H2 react with N2 gives 2 moles of NH3

➙ Then, 5 × 10³ moles = 2/3 × 5 × 10³ moles of NH3

➙ 5 × 10³ moles of H2 = 3.33 × 10³ moles of NH3

We know,

• Molecular weight of NH3 = 17 g
• Moles of NH3 = 3.33 × 10³ moles

By using formula,

➙ Weight of NH3 = 17 × 3.33 × 10³ g

➙ Weight of NH3 = 56.67 × 10³ g

➙ Weight of NH3 = 56.67 kg

✒ Weight of Ammonia formed = 56.67 kg

☀️ Hence, solved !!

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