50.0 L of N2(g) and 10.0 L of H2(g) are mixed at NTP to produce NH3(g). Identify the limiting reagent in the production of NH3 in this situation. Also calculate the volume of NH3 formed at NTP condition. Please anwer both the parts as soon as possible.. I really need help..
Answers
As we know that,
no. of moles=mol. wt.Wt.
Weight of N2=50kg
Molecular weight of N2=28g
No. of moles of N2=2850×103=17.86×102 moles
Weight of H2=10kg
Molecular weight of N2=2g
No. of moles of N2=210×103=5×103 moles
N2+3H2⟶2NH3
From the above reaction,
1 mole of N2 react with 3 moles of H2.
No. of moles of H2 required to react with 17.86×102
moles of N2=3×17.86×102=5.36×103 moles
But only 5×103 moles of H2 are available.
Thus H2 is the limiting reagent here.
Now, again from the above reaction,
Amount of ammonia formed when 3 moles of H2 react =2 moles
Therefore,
Amount of ammonia formed when 5×103 moles of H2
react =32×(5×103)=3.33×103 moles
Molecular weight of ammonia =17g
Weight of ammonia in 3.33×103 moles =17×(3.33×103)=56.61kg.
Hence the amount of NH3 formed is 56.61kg