Respuesta :
N2 + 3H2 --> 2NH3
When 100g of N2 , no of moles of N2= 100/(28)=3.57 mol
no. of moles of h2 = 6/(2)=3mol
Therefore h2 is limiting reagent.
no. of moles of ammonia= 3/3*2=2moles
mass of ammonia produced= 2 mol * (14+3)= 34g
When 100g of N2 , no of moles of N2= 100/(28)=3.57 mol
no. of moles of h2 = 6/(2)=3mol
Therefore h2 is limiting reagent.
no. of moles of ammonia= 3/3*2=2moles
mass of ammonia produced= 2 mol * (14+3)= 34g
Answer : The mass of [tex]NH_3[/tex] produced will be, 34 grams.
Explanation : Given,
Mass of [tex]N_2[/tex] = 100 g
Mass of [tex]H_2[/tex] = 6 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{100g}{28g/mole}=3.57moles[/tex]
[tex]\text{Moles of }H_2=\frac{\text{Mass of }H_2}{\text{Molar mass of }H_2}=\frac{6g}{2g/mole}=3moles[/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, 3 mole of [tex]H_2[/tex] react with 1 mole of [tex]N_2[/tex]
So, the given 3 moles of [tex]N_2[/tex] react with 1 moles of [tex]H_2[/tex]
From this we conclude that, [tex]N_2[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]H_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].
As, 3 moles of [tex]H_2[/tex] react to give 2 moles of [tex]NH_3[/tex]
So, 3 moles of [tex]H_2[/tex] react to give [tex]\frac{3}{3}\times 2=2[/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=(2mole)\times (17g/mole)=34g[/tex]
Therefore, the mass of [tex]NH_3[/tex] produced will be, 34 grams.
