Lithium and nitrogen react in a combination reaction to produce lithium nitride: 6Li (s) + N2 (g) → 2Li3N (s) How many moles of lithium are needed to produce 0.60 mol of Li3N when the reaction is carried out in the presence of excess nitrogen?

12.5 g of Li are needed in order toproduce 0.60 moles of Li₃N

Explanation:

The reaction is:

6Li (s) + N₂ (g) → 2Li₃N (s)

If nitrogen is in excess, the lithium is the limiting reactant.

Ratio is 2:6

2 moles of nitride were produced by 6 moles of Li

Then, 0.6 moles of nitride were produced by (0.6.6) / 2 = 1.8 moles of Li

Let's convert the moles to mass → 1.8 mol. 6.94 g / 1mol = 12.5 g of Li

We need 1.80 moles of lithium

Explanation:

Step 1: Data given

Moles Li3N = 0.60 moles

Step 2: The balanced equation

6Li (s) + N2 (g) → 2Li3N (s)

Step 3: Calculate moles Li

For 2 moles Li3N produced we need 6 moles Li and 1 mol N2

For 0.60 moles Li3N we need 3*0.60 = 1.80 moles Li

We need 1.80 moles of lithium