This is a limiting reagent (LR) problem. You know that because amounts are given for more than one reactant.
You made a typo in the equation. I'm sure you meant to write
N2 + 3H2 = 2NH3
mols N2 = g/molar mass = 25/28 = 0.893
mols H2 = 25/2 = 12.5
mols NH3 produced with 25 g N2 and excess H2 = 0.893 mols N2 x (2 moles NH3/1 mols N2) = 01.78
mols NH3 produced with 25 g H2 and excess N2 = 12.5 mols H2 x (2 mols NH3/3 mols H2) = 8.33
In LR problems the smaller amount always wins since you may only produce the smallest amount of the product. The other reagent is the excess reagent. So N2 is the limiting reagent, you wil produce 0.893 mols NH3 and grams = mols NH3 x molar mass NH3 = ?
Post your work if you get stuck.
If you have 25 g N₂ and 25 g H₂ how many grams of NH₃ could you make?
N2 + 3H3 = 2NH3
1 answer