Final answer:
The percent yield for the reaction forming MgI2, when 3.18 moles of Mg react with 3.56 moles of I2 and 1.76 moles of MgI2 form, is 55.35%.
Step-by-step explanation:
In the reaction Mg (s) + I2 (s) → MgI2 (s), if 3.18 moles of Mg react with 3.56 moles of I2, and 1.76 moles of MgI2 form, the percent yield can be calculated using the actual yield and the theoretical yield. To find the theoretical yield, we need to identify the limiting reactant, which is the reactant that will dictate how much product can be formed. In a stoichiometric reaction, magnesium and iodine would react in a 1:1 mole ratio to form magnesium iodide. Given that there are more moles of iodine than magnesium, magnesium is the limiting reactant. The theoretical yield would be the same as the moles of magnesium since it reacts in a 1:1 ratio with iodine to form magnesium iodide, which means the theoretical yield is 3.18 moles of MgI2.
To find the percent yield, apply the formula:
% Yield = (Actual Yield / Theoretical Yield) × 100
% Yield = (1.76 moles MgI2 / 3.18 moles MgI2) × 100 = 55.35%
Therefore, the percent yield for the formation of MgI2 in this reaction is 55.35%.