What Volume Of 0.150 M FeCl₃(aq) Solution Is Needed To React

What Volume Of 0150 M Fecl3aq Solution Is Needed To React

The assignment asks: "What volume of 0.150 M FeCl3(aq) solution is needed to react completely with 20.0 mL of 0.0450 M AgNO3(aq)? What mass of AgCl will be formed?" The net ionic reaction provided is: Ag+(aq) + Cl-(aq) → AgCl(s). This problem involves stoichiometry and molarity calculations, as well as mass determination based on the limiting reagent principle.

Paper For Above instruction

The given problem revolves around stoichiometric calculations to determine the volume of a ferric chloride solution required to precipitate silver chloride from a given concentration of silver nitrate solution, and subsequently, the mass of AgCl formed. To solve this, several steps are necessary, including writing the relevant reactions, converting concentrations to moles, identifying the limiting reagent, and performing molar ratio calculations.

First, observe the net ionic reaction: Ag+(aq) + Cl-(aq) → AgCl(s). The problem indicates that Ag+ ions originate from AgNO3, while Cl- ions come from FeCl3. Since FeCl3 contains three Cl- ions per formula unit, the molar ratio between FeCl3 and Cl- is 1:3.

The initial concentration of AgNO3 is 0.0450 M, and the volume is 20.0 mL (0.0200 L). Thus, the moles of Ag+ are:

n(Ag+) = concentration × volume = 0.0450 mol/L × 0.0200 L = 9.00 × 10-4 mol.

From the net ionic equation, 1 mol of Ag+ reacts with 1 mol of Cl-. Therefore, the moles of Cl- needed are also 9.00 × 10-4 mol. Since FeCl3 supplies Cl- ions at a ratio of 1:3, the moles of FeCl3 required are:

n(FeCl3) = n(Cl-) / 3 = 9.00 × 10-4 mol / 3 ≈ 3.00 × 10-4 mol.

To find the volume of FeCl3 solution needed, use its concentration:

V(FeCl3) = n / concentration = 3.00 × 10-4 mol / 0.150 mol/L = 2.00 × 10-3 L = 2.00 mL.

Next, to determine the mass of AgCl formed, recognize that all the Ag+ ions precipitate as AgCl, with a molar mass of approximately 143.32 g/mol.

Mass of AgCl = moles × molar mass = 9.00 × 10-4 mol × 143.32 g/mol ≈ 0.129 g.

Thus, 2.00 mL of 0.150 M FeCl3 solution is required to react completely with the silver nitrate, and approximately 0.129 grams of silver chloride will be precipitated in this process.

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