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Calculate the theoretical yield of ammonia produced by the reaction of 100 g of H2 gas and 200g of N2 gas? 3H2(g) + N2(g)-----> 2NH3(g)

User Mangu
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2 Answers

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Answer : The theoretical yield of ammonia produced are, 242.76 g

Solution : Given,

Mass of hydrogen gas = 100 g

Mass of nitrogen gas = 200 g

Molar mass of
H_2 = 2 g/mole

Molar mass of
N_2 = 28 g/mole

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


\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=(100g)/(2g/mole)=50moles


\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=(200g)/(28g/mole)=7.14moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,


3H_2(g)+N_2(g)\rightarrow 2NH_3(g)

From the balanced reaction we conclude that

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

So, 50 moles of
H_2 react with
(50)/(3)=16.66 moles of
N_2

That means, in the given balanced reaction,
N_2 is a limiting reagent because it limits the formation of products and
H_2 is an excess reagent.

Hence, the
N_2 is the limiting reagent.

Now we have to calculate the moles of
NH_3.

As, 1 mole of
N_2 react with 2 moles of
NH_3

So, 7.14 moles of
N_2 react with
2* 7.14=14.28 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=(14.28mole)* (17g/mole)=242.76g

Therefore, the theoretical yield of ammonia produced are, 242.76 g

User Wilhelm Murdoch
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5 votes
Moles of Hydrogen present: 100 / 2 = 50 moles

Moles of Nitrogen present: 200 / 28 = 7.14 moles

Hydrogen required by given amount of nitrogen = 7.14 x 3 = 21.42 moles

Hydrogen is excess so we will calculate the Ammonia produced using Nitrogen.

Molar ratio of Nitrogen : Ammonia = 1 : 2

Moles of ammonia = 7.14 x 2 = 14.28 moles
User Ray Booysen
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7.4k points