Answer : The mass of ammonia produced can be, 121.429 k
Solution : Given,
Mass of
= 100 kg = 100000 g
Mass of
= 100 kg = 100000 g
Molar mass of
= 28 g/mole
Molar mass of
= 2 g/mole
Molar mass of
= 17 g/mole
First we have to calculate the moles of
and
.
![\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7DN_2%3D%5Cfrac%7B%5Ctext%7B%20Mass%20of%20%7DN_2%7D%7B%5Ctext%7B%20Molar%20mass%20of%20%7DN_2%7D%3D%5Cfrac%7B100000g%7D%7B28g%2Fmole%7D%3D3571.43moles)
![\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles](https://tex.z-dn.net/?f=%5Ctext%7B%20Moles%20of%20%7DH_2%3D%5Cfrac%7B%5Ctext%7B%20Mass%20of%20%7DH_2%7D%7B%5Ctext%7B%20Molar%20mass%20of%20%7DH_2%7D%3D%5Cfrac%7B100000g%7D%7B2g%2Fmole%7D%3D50000moles)
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
![N_2+3H_2\rightarrow 2NH_3](https://tex.z-dn.net/?f=N_2%2B3H_2%5Crightarrow%202NH_3)
From the balanced reaction we conclude that
As, 1 mole of
react with 3 mole of ![H_2](https://tex.z-dn.net/?f=H_2)
So, 3571.43 moles of
react with
moles of ![H_2](https://tex.z-dn.net/?f=H_2)
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of ![NH_3](https://tex.z-dn.net/?f=NH_3)
From the reaction, we conclude that
As, 1 mole of
react to give 2 mole of ![NH_3](https://tex.z-dn.net/?f=NH_3)
So, 3571.43 moles of
react to give
moles of ![NH_3](https://tex.z-dn.net/?f=NH_3)
Now we have to calculate the mass of ![NH_3](https://tex.z-dn.net/?f=NH_3)
![\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DNH_3%3D%5Ctext%7B%20Moles%20of%20%7DNH_3%5Ctimes%20%5Ctext%7B%20Molar%20mass%20of%20%7DNH_3)
![\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DNH_3%3D%287142.86moles%29%5Ctimes%20%2817g%2Fmole%29%3D121428.62g%3D121.429kg)
Therefore, the mass of ammonia produced can be, 121.429 kg