If you start with 150.0 L of N2 and an excess of H2 at STP, what volume of NH3 will be made at STP?

Question

N2 (g) + 3H2 (g) > 2NH3 (g)

Answer 1

There are 3 steps to solving this problem

1) Finding how many mols there are in 150L of N₂

2) Finding how many mols of NH₃ are generated

3) Converting how many mols of NH₃ back into Volume.

1) Start with finding out how many mols are in 150 L of N2

Using the ideal gas formula PV=nRT and assuming standard Temperature of 273K, Pressure of 1ATM, and R constant of 0.082

150L*1ATM= n *0.082(L*ATM/K*mol)*273K

150 L*1ATM= n *22.386(L*ATM/mol) Simplify by multiplying K

150 L*1ATM/ 22.386(L*ATM/mol)= n Divide by 22.386 to isolate n.

6.7mols = n

2) Now that we know how many mols are in 150L of N₂, we can look at the given Chemical equation to find out how many mols of N₂ are needed per mol of NH₃

N₂ (g) + 3H₂ (g) > 2NH₃ (g)

We can see here that 2 mols of NH₃ are generated per 1 mol of N₂

(Remember H₂ is excess and won't limit this reaction)

Using that knowledge we know:

6.7mols N₂ --> 13.4NH₃

3) Now to finish by changing mols of NH₃ into Liters.

We'll use the same ideal gas formula but this time we are finding Volume and plugging in mols.

PV=nRT

V *1ATM=13.4mols *0.082(L*ATM/K*mol)*273K

V=300 Liters

Note: you could use intuition to check your answer. Doubling the number of mols and keeping all other variables constant would double the volume.