Question

what is the limiting reactant when 1.50g of lithium and 1.50 g of nitrogen combine to form lithium nitride, a component of advanced batteries, according to the following unbalanced equation?

Answer 1

Limiting reactant is Lithium

Step-by-step explanation:

We are given;

1. Mass of Lithium as 1.50 g
2. Mass of nitrogen is 1.50 g

We are required to determine the rate limiting reagent.

• First, we write the balanced equation for the reaction

6Li(s) + N₂(g) → 2Li₃N

From the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.

• Second, we determine moles of Lithium and nitrogen given.

Moles = Mass ÷ Molar mass

Moles of Lithium

Molar mass of Li = 6.941 g/mol

Moles of Li = 1.50 g ÷ 6.941 g/mol

= 0.216 moles

Moles of nitrogen gas

Molar mass of Nitrogen gas is 28.0 g/mol

Moles of nitrogen gas = 1.50 g ÷ 28.0 g/mol

= 0.054 moles

• According to the equation, 6 moles of Lithium reacts with 1 mole of nitrogen.
• Therefore, 0.216 moles of lithium will require 0.036 moles (0.216 moles ÷6) of nitrogen gas.
• On the other hand, 0.054 moles of nitrogen, would require 0.324 moles of Lithium.

Thus, Lithium is the limiting reagent while nitrogen is in excess.