Questions: In this reaction: Mg(s) + I2(s) -> MgI2(s) If 3.26 moles of Mg react with 3.56 moles of I2, and 1.76 moles of MgI2 form, what is the percent yield?

Solution Steps

To find the limiting reactant, we need to compare the mole ratio of the reactants to the stoichiometric ratio from the balanced chemical equation.

The balanced chemical equation is: \[ \mathrm{Mg}(\mathrm{s}) + \mathrm{I}_2(\mathrm{s}) \rightarrow \mathrm{MgI}_2(\mathrm{s}) \]

From the equation, the mole ratio of Mg to I\(_2\) is 1:1.

Given:

• 3.26 moles of Mg
• 3.56 moles of I\(_2\)

Since the mole ratio is 1:1, the limiting reactant is the one with the fewer moles. Here, Mg has fewer moles (3.26) compared to I\(_2\) (3.56).

Thus, Mg is the limiting reactant.

The theoretical yield is based on the limiting reactant. Since 1 mole of Mg produces 1 mole of MgI\(_2\), 3.26 moles of Mg will produce 3.26 moles of MgI\(_2\).

Theoretical yield of MgI\(_2\) = 3.26 moles

Percent yield is calculated using the formula: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]

Given:

• Actual yield of MgI\(_2\) = 1.76 moles
• Theoretical yield of MgI\(_2\) = 3.26 moles

\[ \text{Percent Yield} = \left( \frac{1.76}{3.26} \right) \times 100 \approx 53.99\% \]

Final Answer