Question

Lithium and nitrogen react to form Li3N, according to the following balanced chemical equation. 6Li + N2 → 2Li3N When 4.0 grams of lithium reacted with 2.5 grams of nitrogen, how much of the non-limiting reactant is left over after the reaction is complete?

Answer 1

6Li + N2 → 2Li3N

no of moles of Li    = W/G.A.Wt

                               = 4/7   = 0.57 moles

no of moles of N2    = W/G..M.Wt

                                = 2.5/28 = 0.0893moles

6 moles of Li react with 1 mole of N2

0.57 moles fo Li react with = 0.57/6   = 0.095moles of N2 is required

N2 is limiting reactant

1 mole of N2 react with 6 mole of Li

0.0893 moles of N2 react with = 6*0.0893/1 = 0.5358 moles of Li is required

Li is excess reactant

The no of moles of excess reactant left over after complete the reaction = 0.57-0.5358   = 0.0342 moles of Li

The mass of excess reactant left over after complete the reaction   = 0.0342*7 = 0.2394g of Li >>>>answer