The equation below shows lithium reacting with nitrogen to produce lithium nitride. 6Li + N2 2Li3N If 12 mol of lithium were reacted with excess nitrogen gas, how many moles of lithium nitride would be produced?

4 moles of Li₃N will be produced in this reaction

Explanation:

The reaction is:

6Li + N₂ → 2Li₃N

If the nitrogen gas is the excess reactant, the limiting must be the lithium.

You always have to make calculations with the limiting reactant. You never use the excess reagent.

Ratio is 6:2.

The rule of three to solve this is:

6 moles of lithium can produce 2 moles of nitride

Therefore, 12 moles of Li must produce (12. 2) / 6 = 4 moles of nitride

4.0 moles of lithium nitride will be produced

Explanation:

Step 1: Data given

Number of moles lithium (Li) = 12.0 moles

Nitrogen gas (N2) is in excess.

Step 2: The balanced equation

6Li + N2 → 2Li3N

Step 3: Calculate moles of lithium nitride (Li3N)

For 6 moles Lithium we need 1 mol nitrogen gas to produce 2 moles lithium nitride

For 12.0 moles lithium we'll have 12.0 / 6 = 2.0 moles nitrogen gas to react to produce 12.0 / 3 = 4.0 moles lithium nitride

4.0 moles of lithium nitride will be produced