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Question

50 kg of N2(g) and 10kg of H2(g) are mixed to produce NH3(g) . Calculate the amount of NH3 formed. Identify the limiting reagent in the production of the NH3.

Solution
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As we know that,
no. of moles=Wt.mol. wt.

Weight of N2=50kg

Molecular weight of N2=28g

No. of moles of N2=50×10328=17.86×102 moles

Weight of H2=10kg

Molecular weight of N2=2g

No. of moles of N2=10×1032=5×103 moles

N2+3H22NH3

From the above reaction,

1 mole of N2 react with 3 moles of H2.

No. of moles of H2 required to react with 17.86×102
moles of N2=3×17.86×102=5.36×103 moles
But only 5×103 moles of H2 are available.
Thus H2 is the limiting reagent here.

Now, again from the above reaction,
Amount of ammonia formed when 3 moles of H2 react =2 moles

Therefore,
Amount of ammonia formed when 5×103 moles of H2
react =23×(5×103)=3.33×103 moles
Molecular weight of ammonia =17g

Weight of ammonia in 3.33×103 moles =17×(3.33×103)=56.61kg.

Hence the amount of NH3 formed is 56.61kg.

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