A lot to do, just to find a percent yield. The first step is to determine just what we might have expected to get if we reacted 12.3 g of lithium with 336 g of nitrogen gas in our own lab. Painful, but this means we need a balanced equation to understand if both reagents are totally consumed, or if one is a limiting reagent, which would reduce the expected yield based on the other reagent.
Lithium nitride has a formula of Li3N.
The balanced equation I come up with is:
Both sides have the same element count. [6 Li, and 2 N]
The balanced equation promises us that 1 mole of N2 gas will react with 6 moles of lithium to form 2 moles of lithium nitride. Let find out how many moles of each material are present in the problem.
The moles of each is obtained by dividing the mass of the material by its molar mass:
Substance Mass(g) Molar mass(g/mole) moles
Li 12.3 6.94 1.77
N2 336 28 12.00
Li3N 5.89 34.8 0.169
We see from the equation that the molar ratio of Li to N2 is 6/1. We need a molar quantity of N2 that is 6x the moles of Li. Since Li is 1.77 moles, 6*(1.77) = 10.6 moles of N2 are required. We have a small excess of N2, so Li is the limiting reagent. One it is consumed, the reaction stops.
Assume all 1.77 moles of lithium react successfully. The equation tells us that we would expect 2 moles Li3N for evey 2 moles Li, a 2/6, or 1/3 molar ratio.
Our yield of Li3N should be (1.77 moles Li)*(1 mole Li3N)/(3 moles Li) = 0.591 moles Li3N
The mass of Li3N is (0.591 moles Li3N)*(34.8 g/mole Li3N) = 20.56 grams Li3N (!)
We only got 5.89 grams. We can quickly conclude one of three possibilities:
1) Our lab partner was sloppy
2) There was an error in the experiment, or
3) Our percent yiled is:
(5.89 g actual)/(20.56 g theorectical) = 0.286 or 28.6%
