5o kg of N2 AND 10.0 KG OF H2 ARE MIXED TO PRODUCE NH3. CALCULATE THE NH3 FORMED. IDENTIFY THE LIMITING REAGENT IN THE PRODUCTION OF NH3 IN THIS SITUATION.

Asked by  | 10th Jul, 2013, 08:50: PM

Expert Answer:

 

N2 + 3H2 ? 2NH3.

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

Molecular mass of Hydrogen = 6 g/mol = 0.006 kg/mol

Molecular mass of Ammonia = 17 g/mol = 0.017 kg/mol

Now, according to the balanced chemical equation,

0.028 kg of Nitrogen reacts with 0.006 kg of Hydrogen.

So, 50 kg of Nitrogen reacts with (0.006 x 50) / 0.028  = 10.71 kg of Hydrogen.

The amount of Hydrogen given (10 kg) is less than the amount required (10.71 kg) for 50 kg of Nitrogen. Therefore, Hydrogen is the limiting reagent. the formation of Ammonia will depend on the amount of Hydrogen  available for reaction.

0.006 kg of Nitrogen produces (2x0.017) = 0.034 kg of Ammonia.

So, 10 kg of Hydrogen will produce (0.034 x 10) / 0.006 = 56.67 kg of Ammonia.

Answered by  | 10th Jul, 2013, 10:46: PM

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