Question

What amount of the excess reagent remains when 0.500 mol Li reacts with 0.350 mol N₂?

6Li(s) + N₂(g) rightarrow 2Li₃ N(s)

(A) 0.0833 mol N₂
(B) 0.150 mol Li
(C) 0.267 mol N₂
(D) 0.417 mol Li

Answer 1

When 0.500 mol of lithium reacts with 0.350 mol of nitrogen, lithium is the limiting reactant, and 0.267 mol of nitrogen gas remains as the excess reagent after the reaction is complete.

Step-by-step explanation:

To determine the amount of excess reagent remaining when 0.500 mol Li reacts with 0.350 mol N₂, we need to use the stoichiometry of the balanced chemical equation:

6Li(s) + N₂(g) → 2Li₃N(s)

From the equation, every 6 moles of lithium react with 1 mole of nitrogen gas to form 2 moles of lithium nitride. To find out which reagent is limiting, we compare the mole ratio of the reactants provided with the mole ratio in the balanced equation.

The mole ratio of Li to N₂ is 6:1, which means for every 6 moles of Li, we need 1 mole of N₂. We have 0.500 mol Li, so the corresponding amount of N₂ required is:

0.500 mol Li × (1 mol N₂ / 6 mol Li) = 0.0833 mol N₂

Since we have 0.350 mol N₂, which is more than 0.0833 mol, it means that N₂ is in excess and Li is the limiting reactant.

The amount of excess N₂ can be calculated by:

0.350 mol N₂ - 0.0833 mol N₂ = 0.267 mol N₂

Hence, the correct answer is (C) 0.267 mol N₂ remains as the excess reagent.