The amount of the excess reagent remains when 0.500 mole of Li reacts with 0.350 mole N₂ is 0.267 mole
How to calculate the amount of excess reagent remaining?
It important that we determine the excess reagent if we are to calculate the amount of the excess reagent that remains.
The excess reagent is determined as follow:
From the balanced equation,
6 moles of Li reacted with 1 mole of N₂
Therefore,
0.5 mole of Li will react with = 0.5 / 6 = 0.083 mole of N₂
Since 0.083 mole of N₂, thus, N₂ is the excess reagent.
Finally, we shall calculate the amount of the excess reagent that remains. This is shown below:
- Mole of excess reagent before reaction = 0.350 mole
- Mole of excess reagent that reacted = 0.083 mole
- Mole of excess reagent that remains =?
Mole of excess reagent that remains = Mole before - mole that reacts
= 0.350 - 0.083
= 0.267 mole
Complete question:
What amount of the excess reagent remains when 0.500 mol Li reacts with 0.350 mol N2?