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1 vote
If you feed 100 kg of N2 gas and 100 kg of H2 gas into a

reactor: What is the maximum number of kg of NH3 that can be
produced?

User Auhuman
by
8.3k points

1 Answer

4 votes

Answer : The mass of ammonia produced can be, 121.429 k

Solution : Given,

Mass of
N_2 = 100 kg = 100000 g

Mass of
H_2 = 100 kg = 100000 g

Molar mass of
N_2 = 28 g/mole

Molar mass of
H_2 = 2 g/mole

Molar mass of
NH_3 = 17 g/mole

First we have to calculate the moles of
N_2 and
H_2.


\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=(100000g)/(28g/mole)=3571.43moles


\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=(100000g)/(2g/mole)=50000moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,


N_2+3H_2\rightarrow 2NH_3

From the balanced reaction we conclude that

As, 1 mole of
N_2 react with 3 mole of
H_2

So, 3571.43 moles of
N_2 react with
3571.43* 3=10714.29 moles of
H_2

From this we conclude that,
H_2 is an excess reagent because the given moles are greater than the required moles and
N_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
NH_3

From the reaction, we conclude that

As, 1 mole of
N_2 react to give 2 mole of
NH_3

So, 3571.43 moles of
N_2 react to give
3571.43* 2=7142.86 moles of
NH_3

Now we have to calculate the mass of
NH_3


\text{ Mass of }NH_3=\text{ Moles of }NH_3* \text{ Molar mass of }NH_3


\text{ Mass of }NH_3=(7142.86moles)* (17g/mole)=121428.62g=121.429kg

Therefore, the mass of ammonia produced can be, 121.429 kg

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