In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 2.34 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what is the percent yield?

Answer 1

Since the molar ratio between the Mg & MgI₂ is 1:1, (as is that between the I₂ & the MgI₂) and we see that the moles of Mg reacted is less than that of the iodine. Hence, the Mg is the limiting reactant and the theoretical yield of MgI₂ would be 2.34 moles.

The actual yield is given as 1.76 moles. So the % Yield would be found as follows:

(AY/TY) x 100 = 7.52% (to 3 sig figs)