Question

Calcium metal will react with iodine to produce solid calcium iodide according to the unbalanced equation: Ca(s) + I₂(s) → CaI₂(s). If four moles of calcium metal and five moles of iodine are used, how many moles of calcium iodide could be produced in this reaction?

Answer 1

Using the balanced chemical equation, Ca(s) + I2(s) → CaI2(s), with 4 moles of calcium and 5 moles of iodine, up to 4 moles of calcium iodide could be produced because calcium is the limiting reactant.

Step-by-step explanation:

The stoichiometry of the reaction between calcium metal and iodine to produce calcium iodide can be understood by balancing the equation and determining the limiting reactant. The balanced chemical equation is Ca(s) + I2(s) → CaI2(s). When we have 4 moles of calcium and 5 moles of iodine, we can predict the amount of product formed based on the ratios dictated by the balanced equation.

According to the equation, 1 mole of calcium reacts with 1 mole of iodine to produce 1 mole of calcium iodide. Here, calcium is the limiting reactant because we have excess iodine (we only need 4 moles of iodine to react with 4 moles of calcium), and the reaction will stop once all calcium is consumed. Therefore, we could produce up to 4 moles of calcium iodide.