Ammonia burns in air as follows: 4NH3(g) + 3O2(g) ? 2N2(g) + 6H2O(g)

Question

I am trying to figure out this question, but I have no idea how. Please help!

 

Ammonia burns in air as follows: 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(g), How many liters of N2(g) are formed by burning 5.38 L of ammonia and 5.38 L of oxygen with all species at the same temperature and pressure?

Answer 1

Since all species is at the same temperature and pressure, this problem can be simply determined by first identifying the limiting reagent, then using the fact that volume is directly proportional to moles.

 

Since there requires more ammonia than oxygen in the reaction, we can see that for equal volumes, the ammonia will limit the reaction. Thus, we have 5.38L of ammonia reacting. Then, we use the fact that for every 2 moles of ammonia, we generate 1 mole of nitrogen, thus, we halve the number of moles of ammonia reacted to find the number of moles of Nitrogen. Since moles are directly proportional to volume, we halve the volume of ammonia to determine the volume of nitrogen.

 

Therefore, there is 2.69L of nitrogen formed.