If 2 mol Li solid reactants with 1.5 mole N₂ gas, how many moles of Li₃N solid forms? 6Li (s) + N₂(g) --> 2Li₃N (s)?
If 2 mol Li solid reactants with 1.5 mole N₂ gas, how many moles of Li₃N solid forms?
6Li (s) + N₂(g) --> 2Li₃N (s)
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2 moles of Li would react completely with 2 x (1/6) = 1/3 mole of N2, but there is more N2 present than that, so N2 is in excess and Li is the limiting reactant.
(2 mol Li) x (2 mol Li₃N / 6 mol Li) = 2/3 mol Li₃N