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)

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

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BarrettArcher BarrettArcher · Answer 2 · 2019-02-25T00:45:08-05:00

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