When calcium is allowed to react with nitrogen, N2, calcium nitride is formed, as shown in the following balanced chemical equation: 3Ca + N2 mc006-1.jpg Ca3N2

Question

If 24.0 g of calcium and 12.0 g of nitrogen are available for this reaction, the limiting reagent will be

Answer 1

I work these the long way and not the short way.
3Ca + N2 ==> Ca3N2
mols Ca = g/atomic mass = 24.0/40 = about 0.6
mols N2 = 12/28 = about 0.43
Take them one at a time.
Moles Ca3N2 formed using Ca it is 0.6 x (1 mol Ca3N2/3 mols Ca) = 0.6 x 1/3 = 0.2.
mols Ca3N2 formed using N2 it is 0.43 x (1 mol Ca3N2/1 mol N2) = 0.43
0.2 is the smaller number; therefore, Ca is the limiting reagent (LR)