Answer:
Complete reaction
mole %: 0% N₂; 20% H₂; 80% NH₃
Gram %: 0% N₂; 2,9% H₂; 97,1% NH₃
75% reaction:
mole%: 8,3% N₂; 41,7% H₂; 50% NH₃
Gram: 20% N₂; 7,2% H₂; 72,8% NH₃
Explanation:
The global reaction in the reactor is:
N₂(g) + 3 H₂ (g) → 2 NH₃ (g)
For a complete reaction of 20 moles of N₂(g) you need:
20 moles N₂ ₓ [tex]\frac{3 mol H_2}{1 mol N_2}[/tex] = 60 moles of H₂(g)
So, 10 moles of H₂(g) will stay in the end.
The moles produced of NH₃ are:
20 moles N₂ ₓ [tex]\frac{2 mol NH_3}{1 mol N_2}[/tex] = 40 moles of NH₃(g)
Thus, the final composition in moles is:
0 moles N₂; 10 moles H₂; 40 moles of NH₃
The mole % is:
0% N₂; ¹⁰/₅₀ ₓ 100 = 20% H₂; ⁴⁰/₅₀ ₓ100 = 80% NH₃
In grams you have:
10 moles H₂ [tex]\frac{2,02 g}{1 mol}[/tex] = 20,2 g of H₂
40 moles NH₃ [tex]\frac{17,03 g}{1 mol}[/tex] = 681,2 g of NH₃
0% N₂; [tex]\frac{20,2 g}{701,4}[/tex] ₓ 100 = 2,9% H₂; [tex]\frac{681,2 g}{701,4}[/tex] ₓ100 = 97,1% NH₃
With a 75% of conversion you have that the moles produced of NH₃ are:
40 moles of NH₃(g) × 75% = 30 moles of NH₃
The necessary moles to produce these moles of NH₃ are:
30 moles NH₃ ₓ [tex]\frac{3 mol H_2}{2 mol NH_3}[/tex] = 45 moles of H₂(g)
So, 25 moles of H₂(g) will stay in the end.
30 moles NH₃ ₓ [tex]\frac{1 mol N_2}{2 mol NH_3}[/tex] = 15 moles of N₂(g)
So, 5 moles of H₂(g) will stay in the end.
Thus, the final composition in moles is:
5 moles N₂; 25 moles H₂; 30 moles of NH₃
The mole % is:
⁵/₆₀ ₓ 100 =8,3% N₂; ²⁵/₆₀ ₓ 100 = 41,7% H₂; ³⁰/₆₀ ₓ100 = 50% NH₃
In grams you have:
5 moles N₂ [tex]\frac{28 g}{1 mol}[/tex] = 140 g of N₂
25 moles H₂ [tex]\frac{2,02 g}{1 mol}[/tex] = 50,5 g of H₂
30 moles NH₃ [tex]\frac{17,03 g}{1 mol}[/tex] = 510,9 g of NH₃
[tex]\frac{140 g}{701,4}[/tex] ₓ 100 =20% N₂; [tex]\frac{ 50,5g}{701,4}[/tex] ₓ 100 = 7,2% H₂; [tex]\frac{510,9 g}{701,4}[/tex] ₓ100 = 72,8% NH₃
I hope it helps!
