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Nuetrik [128]
3 years ago
10

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)-----&g

t; 2NH3(g)
Chemistry
2 answers:
Alex787 [66]3 years ago
8 0
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
pashok25 [27]3 years ago
3 0

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}=\frac{100g}{2g/mole}=50moles

\text{Moles of }N_2=\frac{\text{Mass of }N_2}{\text{Molar mass of }N_2}=\frac{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 \frac{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\times 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\times \text{Molar mass of }NH_3

\text{Mass of }NH_3=(14.28mole)\times (17g/mole)=242.76g

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

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