Question

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?

Answer 1

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