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?

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

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