Question:

N₂(g) + 3 H₂(g) →→→ 2NH₂(g) Assume 0.220 mol N₂ and 0.704

Last updated: 7/31/2022

N₂(g) + 3 H₂(g) →→→ 2NH₂(g) Assume 0.220 mol N₂ and 0.704

N₂(g) + 3 H₂(g) →→→ 2NH₂(g) Assume 0.220 mol N₂ and 0.704 mol H₂ are present initially. After complete reaction, how many moles of ammonia are produced? How many moles of H₂ remain? How many moles of N₂ remain?