Respuesta :
Answer : The mass of ammonia produced can be, 121.429 k
Solution : Given,
Mass of [tex]N_2[/tex] = 100 kg = 100000 g
Mass of [tex]H_2[/tex] = 100 kg = 100000 g
Molar mass of [tex]N_2[/tex] = 28 g/mole
Molar mass of [tex]H_2[/tex] = 2 g/mole
Molar mass of [tex]NH_3[/tex] = 17 g/mole
First we have to calculate the moles of [tex]N_2[/tex] and [tex]H_2[/tex].
[tex]\text{ Moles of }N_2=\frac{\text{ Mass of }N_2}{\text{ Molar mass of }N_2}=\frac{100000g}{28g/mole}=3571.43moles[/tex]
[tex]\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{100000g}{2g/mole}=50000moles[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]N_2[/tex] react with 3 mole of [tex]H_2[/tex]
So, 3571.43 moles of [tex]N_2[/tex] react with [tex]3571.43\times 3=10714.29[/tex] moles of [tex]H_2[/tex]
From this we conclude that, [tex]H_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]N_2[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]NH_3[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]N_2[/tex] react to give 2 mole of [tex]NH_3[/tex]
So, 3571.43 moles of [tex]N_2[/tex] react to give [tex]3571.43\times 2=7142.86[/tex] moles of [tex]NH_3[/tex]
Now we have to calculate the mass of [tex]NH_3[/tex]
[tex]\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3[/tex]
[tex]\text{ Mass of }NH_3=(7142.86moles)\times (17g/mole)=121428.62g=121.429kg[/tex]
Therefore, the mass of ammonia produced can be, 121.429 kg
