How many grams of Li3N can be formed from 1.75 moles of Li? Assume an excess of nitrogen. 6 Li (s) + N2 (

g. → 2 Li3N (s)

The balance chemical reaction is:

6 Li (s) + N2 (g) → 2 Li3N (s)

We use the amount of the Lithium reactant and the relations of the substances from the reaction to calculate the amount of product produced. We do as follows:

1.75 mol Li (2 mol Li3N / 6 mol Li) (34.83 g / mol) = 20.32 g Li