I A \(50.00-\mathrm{mL}\) sample of \(0.0250 \mathrm{M}\) silver nitrate is mixed with \(0.0400 \mathrm{M}\) chromium(III) chloride. (a) What is the minimum volume of chromium(III) chloride required to completely precipitate silver chloride? (b) How many grams of silver chloride are produced from (a)?

Balancing chemical equations is a fundamental skill in chemistry, involving the adjustment of the coefficients in front of reactants and products to ensure the conservation of matter. It's essential that the quantities of atoms for each element are the same on both sides of the equation. In the case of silver nitrate reacting with chromium(III) chloride to form silver chloride and chromium nitrate, we observe a 3:1 mole ratio. This means you need three silver ions for every one chromium ion to keep the balance, as expressed in the equation:

\(3 \text{AgNO}_3 (aq) + \text{CrCl}_3 (aq) \rightarrow 3 \text{AgCl} (s) + \text{Cr(NO}_3\text{)}_3 (aq)\).

Understanding the 3:1 ratio is vital because it will be used to conclude how much of each reactant is needed, leading us to the concept of stoichiometry, which is utilized in later calculations.

Molarity is a measure of concentration referring to the number of moles of a compound per liter of solution. This calculation is important when mixing solutions to ensure that reactions proceed correctly. To find the moles of silver nitrate (\(\text{AgNO}_3\)), one would multiply the volume of the solution by its molarity, here:

\(\text{Moles of AgNO}_3 = (50.00 \text{ mL} \times 0.0250 \text{ mol/L}) = 1.25 \times 10^{-3} \text{ mol}\).

The molarity concept helps us understand the relationship between the volume of chromium(III) chloride needed and its concentration. Moreover, this is directly linked to the stoichiometric ratio from the balanced chemical equation. Knowing one's way around molarity calculations is crucial for determining how different concentrations of reactants affect the amount of product formed.

A precipitation reaction occurs when two aqueous solutions combine to form a precipitate, which is an insoluble solid. In our exercise, when silver nitrate and chromium(III) chloride are mixed, silver chloride, a precipitate, forms. The balanced equation tells us that three moles of silver nitrate will produce three moles of silver chloride, hence the stoichiometric relationship used in the calculation. To calculate the mass of silver chloride precipitated, one must use the molar mass (the sum of the atomic masses of all atoms in the molecule) and the moles we've already determined:

\(\text{Mass of AgCl} = (1.25 \times 10^{-3} \text{ mol}) \times (143.32 \text{ g/mol}) \approx 0.179 \text{ g}\).

Understanding precipitation reactions is key in fields like analytical chemistry for separating components or in environmental chemistry for removing pollutants from water.