Question

When heated lithium reacts with nitrogen to form lithium nitride: 6Li(s) + N2(g) -----> 2Li3N(S) What...

When heated lithium reacts with nitrogen to form lithium nitride: 6Li(s) + N2(g) -----> 2Li3N(S)

What is the theoratrical yield of Li2N in grams when 12.9 of Li is heated with 35.2 g of N2?

If the actual yield of Li3N is 6.49g what is the % yield of the reaction?

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Answer #1

actual yield = 6.49 grams Li3N

grams N2 required = (12.9 grams Li) * (1 mol Li / 6.941 grams Li) * (1 mol N2 / 6 mol Li) * (28 grams N2 / 1 mol N2) = 8.673 grams N2 required

Since we have enough N2, then Li is the limiting reactant

theoretical yield Li3N = (12.9 grams Li) * (1 mol Li / 6.941 grams Li) * (2 mol Li3N / 6 mol Li) * (34.823 grams Li3N / 1 mol Li3N) = 21.57 grams Li3N produced

percent yield = (actual yield / theoretical yield) * 100% = (6.49 grams Li3N / 21.57 grams Li3N) * 100% = 30.088 % is the answer.

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