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Chemical reactions are processes in which substances, known as reactants, transform into new substances, called products. In the laboratory preparation of iodine, chlorine gas (\( \mathrm{Cl}_{2}(g) \)) reacts with sodium iodide solution (\( \mathrm{NaI}(aq) \)). The chemical equation for this reaction is already balanced:

• \( 2 \mathrm{NaI}(aq) + \mathrm{Cl}_{2}(g) \rightarrow \mathrm{I}_{2}(s) + 2 \mathrm{NaCl}(aq) \)

A balanced equation is crucial because it ensures the conservation of mass and moles, highlighting the stoichiometry of the reaction.
This stoichiometry tells us the relationship between the numbers of moles of each reactant and product involved. For example, 2 moles of sodium iodide produce 1 mole of iodine. This stoichiometric ratio helps in understanding the quantities of reactants needed to form a desired amount of product.

Molar mass is an essential concept in stoichiometry. It helps us convert between the mass of a substance and the moles of that substance. Molar mass is the mass of one mole of a substance, usually expressed in grams per mole. For example, the molar mass of diatomic iodine (\( \mathrm{I}_{2} \)) is calculated by adding the atomic masses of the iodine atoms:

• Atomic mass of iodine: approximately 126.90 g/mol
• Molar mass of \( \mathrm{I}_{2} \) = 2 x 126.90 = 253.80 g/mol

Understanding molar masses allows us to understand how much of one chemical is equivalent to one mole, which is critical when calculating how much of a reactant is needed or product is formed in a reaction.Similarly, calculating the molar mass of sodium iodide (\( \mathrm{NaI} \)) involves adding the atomic masses of sodium and iodine, which results in a molar mass of approximately 149.89 g/mol.

Mole calculations are a key part of stoichiometry and involve converting between mass, moles, and (sometimes) number of molecules. These calculations are grounded in the concept of the mole—a central unit in chemistry that links the macroscopic and atomic scales.For instance, when you have 50.0 grams of iodine, you can find how many moles of iodine are present by using its molar mass:

• Moles of \( \mathrm{I}_{2} \) = \( \frac{50.0 \text{ g}}{253.80 \text{ g/mol}} \approx 0.197 \text{ mol} \)

Once you know the moles of iodine, you can use their stoichiometry to find the moles of sodium iodide needed. Since the reaction requires 2 moles of sodium iodide to produce 1 mole of iodine, you can determine the moles of sodium iodide:

• Moles of \( \mathrm{NaI} \) = 2 x 0.197 mol = 0.394 mol

Finally, to find out how much sodium iodide mass is required, multiply the moles of sodium iodide by its molar mass:

• Mass of \( \mathrm{NaI} \) = 0.394 mol x 149.89 g/mol \( \approx 59.06 \text{ g} \)

These calculations show the practical application of stoichiometry in predicting the amounts of substances needed or generated in chemical reactions.