In this reaction: Mg (s) + I₂ (s) → MgI₂ (s), if 10.0 g of Mg reacts with 60.0 g of I₂, and 60.68 g of MgI₂ form, what is the percent yield?

Answer 1

Mg(s) + I₂(s) ==> MgI₂(s) ... balanced equation

First, find the limiting reactant.

moles Mg present = 10.0 g Mg x 1 mol Mg/24.3 g = 0.412 mols Mg

moles I2 present = 60.0 g I₂ x 1 mol I2/254 g = 0.236 mols I₂

Since mol ratio of Mg:I₂ in balanced equation is 1:1, limiting reactant = I₂

To find % yield, we must first calculate the theoretical yield:

Theoretical yield = 0.236 mols I2 x 1 mol MgI2 / 1 mol I2 = 0.236 mols MgI2 x 278 g MgI2/mol = 65.7 g

% yield = actual yield / theoretical yield (x100%) = 60.68 g / 65.7 g (x100%) = 92.4% yield (3 sig figs)