Question

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)

Answer 1

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

Answer 2

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