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?

  98.9%

Explanation:

2 moles of I₂ are required for each mole of Mg, so the reaction is limited by the available I₂. The 3.56 moles of I₂ should react with 1.78 moles of Mg to produce 1.78 moles of MgI₂. Instead, we get 1.76 moles of MgI₂.

The yield is 1.76/1.78 × 100% ≈ 98.876%

The yield is 98.9% of the quantity expected based on available reactants.

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