Question

Lithium and nitrogen react in a combination reaction to produce lithium nitride:

6Li(s) + N₂(g) → 2Li₃N(s)

In a particular experiment, 3.00 g samples of each reagent are reacted. The theoretical yield of lithium nitride is ________ g.

Answer 1

6 Li(s) +N2(g) $\rightarrow$ 2Li3N(s)

number of moles of Li= amount of Li (g)/ molar mass of Li= 3 g/ 6.94 g/mol= 0.4323 moles

number of moles of N2= amount of N2 (g)/ molar mass of N2= 3 g/ 28.01 g/mol= 0.1071 moles

According to balance equation

6 moles of Li requires 1 mole of N2

therefore 0.4323 moles of Li require will require=   1/6 X 0.4323 mole of N2= 0.0721 moles of N2

hence N2 is excess reagent

According to balance equation

6 moles of Li produces 2 moles of Li3N

therefore

0.4323 moles of Li will produces= 2/6 X 0.4323 moles of Li3N = 0.1441 moles of Li3N

Amount of Li3N produced in g= number of moles of Li3N produced X mw of Li3N (g/mol)

= 0.1441 mole X 34.82 g/mole = 5.02 g answer