Question

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)?

Short answer

Answer: The minimum volume of chromium(III) chloride required is 15.63 mL, and the mass of silver chloride produced is 0.17915 grams.

Step-by-step solution

Write the balanced chemical equation

First, we need to write the balanced chemical equation for the reaction between silver nitrate (AgNO3) and chromium(III) chloride (CrCl3) to form silver chloride (AgCl) and chromium(III) nitrate (Cr(NO3)3): 2 AgNO3(aq) + CrCl3(aq) ➞ 2 AgCl(s) + Cr(NO3)3(aq)

Determine moles of silver nitrate

Since we are given the volume and concentration of the silver nitrate solution, we can determine the moles of silver nitrate present using the formula: moles = volume × concentration moles of AgNO3 = (50.00 mL) × (0.0250 mol/L) Note that 1 L = 1000 mL, so we must convert the volume to liters: moles of AgNO3 = (50.00 × 10^{-3} L) × (0.0250 mol/L) = 1.250 × 10^{-3} mol

Calculate required moles of chromium(III) chloride

From the balanced chemical equation, we see that the stoichiometric ratio between AgNO3 and CrCl3 is 2:1. To completely react with all the silver nitrate, we will need half the moles of chromium(III) chloride: moles of CrCl3 = (1.250 × 10^{-3} mol) / 2 = 6.250 × 10^{-4} mol

Calculate required volume of chromium(III) chloride

Now that we have the moles of chromium(III) chloride required, we can calculate the volume of the solution needed at the given concentration: volume = moles ÷ concentration minimum volume of CrCl3 = (6.250 × 10^{-4} mol) ÷ (0.0400 mol/L) = 1.563 × 10^{-2} L Convert the volume to milliliters: minimum volume of CrCl3 = (1.563 × 10^{-2} L) × (1000 mL/L) = 15.63 mL

Calculate mass of silver chloride produced

The stoichiometric ratio between AgNO3 and AgCl in the balanced chemical equation is 1:1. Therefore, the moles of AgCl produced will be equal to the moles of AgNO3: moles of AgCl = moles of AgNO3 = 1.250 × 10^{-3} mol The molar mass of AgCl is 143.32 g/mol. We can now calculate the mass of AgCl produced: mass of AgCl = moles of AgCl × molar mass of AgCl mass of AgCl = (1.250 × 10^{-3} mol) × (143.32 g/mol) = 0.17915 g (a) The minimum volume of chromium(III) chloride required to completely precipitate silver chloride is 15.63 mL. (b) The mass of silver chloride produced is 0.17915 grams.

Key concepts

Stoichiometry in Precipitation Reactions

Stoichiometry is the mathematics behind chemistry. It allows us to predict the outcomes of chemical reactions, calculating how much of each reactant is needed to react completely with the others, and what quantities of products will be generated. For instance, in precipitation reactions—one of the classical types of reactions—we mix two solutions, leading to the formation of an insoluble compound that precipitates, or settles out of the solution.

To understand how this applies to our textbook exercise, consider the reaction of silver nitrate with chromium(III) chloride which produces silver chloride. We start by writing a balanced chemical equation. Balancing tells us the ratio of reactants that react together; in this case, two moles of silver nitrate react with one mole of chromium(III) chloride, forming silver chloride. This stoichiometric ratio is crucial, as it dictates that for every mole of chromium(III) chloride, we need twice as many moles of silver nitrate to fully react.

By comparing the stoichiometric coefficients of the reactants and products in the balanced equation, we can calculate the exact amounts of reactants needed to produce a desired amount of product, guiding us to the answer for both parts (a) and (b) of our exercise.

Molar Mass Calculation

Molar mass represents the weight of one mole of a substance, commonly expressed in grams per mole (g/mol). This concept ties closely to stoichiometry when converting between mass of a substance and the amount in moles. For chemical reactions, molar mass serves as a conversion factor that can bridge moles to grams and vice versa.

In the example of forming silver chloride (AgCl), knowing its molar mass is critical to find out how many grams can be produced from the given moles of silver nitrate. For AgCl, with one silver atom (Ag) and one chlorine atom (Cl), we calculate its molar mass by adding together the molar masses of silver and chlorine, round about 107.87 g/mol and 35.45 g/mol, respectively. This gives us a molar mass for AgCl of approximately 143.32 g/mol.

With this number available, we multiply the moles of AgCl by its molar mass to find the mass in grams, a crucial step for solving part (b) of the exercise. This conversion is a key aspect of stoichiometry and essential for quantitative analysis in chemistry.

Convert Volume to Moles

Converting volume to moles is a common practice in chemistry when dealing with solutions. Concentration, typically expressed in moles per liter (M or mol/L), provides a direct way to perform this conversion, as long as we clearly understand the relationship between the volume of the solution and its concentration.

In our exercise, we're given the volume in milliliters and concentration in molarity (M) of the silver nitrate solution to calculate moles. To do this accurately, it's important to convert the volume from milliliters to liters since the concentration is given in terms of liters. This is done by recognizing that 1 L of a solution contains 1000 mL. Then, by multiplying the volume (now in liters) by the concentration (mol/L), we get the number of moles of the substance in that volume of solution.

This process enables us to tackle part (a) of the exercise, finding the minimum volume of chromium(III) chloride needed to cause complete precipitation with the available silver nitrate. It is these very principles of converting volume to moles that allow chemists to scale reactions up or down and to prepare solutions with precise concentrations for laboratory experiments and industrial processes alike.