Question

Consider a 1.50-g mixture of magnesium nitrate and magnesium chloride. After dissolving this mixture in water, 0.500 M silver nitrate is added dropwise until precipitate formation is complete. The mass of the white precipitate formed is 0.641 g. a. Calculate the mass percent of magnesium chloride in the mixture. b. Determine the minimum volume of silver nitrate that must have been added to ensure complete formation of the precipitate.

Short answer

a) The mass percent of magnesium chloride in the mixture is 14.21%. b) The minimum volume of silver nitrate that must have been added to ensure complete formation of the precipitate is 8.952 mL.

Step-by-step solution

Write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction between magnesium chloride and silver nitrate is: \[ MgCl_{2(aq)} + 2 AgNO_{3(aq)} → 2 AgCl_{(s)} + Mg(NO_{3})_{2(aq)} \] This equation tells us that 1 mole of magnesium chloride reacts with 2 moles of silver nitrate to form 2 moles of silver chloride.

Calculate the moles of silver chloride formed

We are given the mass of silver chloride precipitate (0.641 g) and we can find the molar mass of silver chloride to be: M(AgCl) = 107.87 (Ag) + 35.45 (Cl) = 143.32 g/mol Now, we can calculate the moles of silver chloride formed: moles of AgCl = mass / molar mass moles of AgCl = 0.641 g / 143.32 g/mol = 0.004476 mol

Calculate the moles of magnesium chloride in the mixture

From the balanced chemical equation, we know that 2 moles of silver chloride come from 1 mole of magnesium chloride. Thus, we can find the moles of magnesium chloride: moles of MgCl2 = (moles of AgCl) / 2 = 0.004476 / 2 = 0.002238 mol

Calculate the mass of magnesium chloride in the mixture

We can find the molar mass of magnesium chloride as follows: M(MgCl2) = 24.31 (Mg) + 2 × 35.45 (Cl) = 95.21 g/mol Now, we can calculate the mass of magnesium chloride in the mixture: mass of MgCl2 = moles × molar mass mass of MgCl2 = 0.002238 mol × 95.21 g/mol = 0.2131 g

Calculate the mass percent of magnesium chloride in the mixture

We are given the total mass of the mixture (1.50 g), and we have calculated the mass of magnesium chloride in the mixture (0.2131 g). The mass percent of magnesium chloride in the mixture can be calculated as: mass percent of MgCl2 = (mass of MgCl2 / total mass of mixture) × 100% mass percent of MgCl2 = (0.2131 g / 1.50 g) × 100% = 14.21 % a) The mass percent of magnesium chloride in the mixture is 14.21%.

Calculate the moles of silver nitrate needed for the reaction.

From the balanced chemical equation, we know that 1 mole of magnesium chloride reacts with 2 moles of silver nitrate (a 1:2 ratio). So, we can find the moles of silver nitrate needed for the reaction: moles of AgNO3 = 2 × moles of MgCl2 = 2 × 0.002238 mol = 0.004476 mol

Calculate the volume of 0.500 M silver nitrate solution needed

We are given the concentration of silver nitrate (0.500 M) and we have just calculated the moles needed for the reaction (0.004476 mol). We can now find the volume of silver nitrate needed: volume = moles / concentration volume = 0.004476 mol / 0.500 M = 0.008952 L Since we want the answer in milliliters, we can convert liters to milliliters: volume = 0.008952 L × 1000 mL/L = 8.952 mL b) The minimum volume of silver nitrate that must have been added to ensure complete formation of the precipitate is 8.952 mL.

Key concepts

Understanding Precipitation Reactions

Precipitation reactions are a type of chemical reaction in which two soluble salts in aqueous solutions react to form an insoluble salt or precipitate. This occurs when the cation from one compound reacts with the anion from another compound, forming a solid. For example, in the reaction between magnesium chloride and silver nitrate, silver chloride is the precipitate that forms: - The equation for this reaction is: \[ MgCl_{2(aq)} + 2 AgNO_{3(aq)} \rightarrow 2 AgCl_{(s)} + Mg(NO_{3})_{2(aq)} \] - Silver chloride \((AgCl)\) appears as a solid white precipitate. Precipitation reactions are often used to separate and identify components in a mixture. Recognizing when and how these reactions occur forms a basic yet essential concept in chemical stoichiometry.

Mass Percent Calculation

The mass percent of a component in a mixture reveals its proportion relative to the total mass. It's a percent weight calculation, useful in understanding compositions of substances. This is calculated by dividing the mass of the individual component by the total mass of the mixture and then multiplying by 100%.To determine this in our example: - We found 0.2131 grams of magnesium chloride in a 1.50-gram mixture. - By dividing \(0.2131 \text{ g} \) by \( 1.50 \text{ g} \) and multiplying by 100%, we find a mass percent of approximately 14.21% for magnesium chloride.This percentage helps in quantifying exactly how much magnesium chloride was in the initial mixture, crucial for precise chemical analysis.

The Mole Concept

The mole is a fundamental unit in chemistry that allows chemists to count particles at the atomic scale, connecting macroscopic quantities with microscopic particles. It enables you to convert between atoms, molecules, and grams via the molar mass.Applying this to our exercise: - We calculated the moles of silver chloride using its mass \((0.641 \text{ g})\) and molar mass \((143.32 \text{ g/mol})\). Thus, moles of \(AgCl = \frac{0.641}{143.32} = 0.004476 \text{ mol}\). - From the balanced equation, these moles relate directly to the moles of magnesium chloride \((MgCl_2)\), thanks to stoichiometry.The mole concept allows accurate chemical calculations to ensure reaction equivalency and ingredient proportions.

Understanding Solution Concentration

Solution concentration expresses the amount of solute present in a specific volume of solvent. It’s often expressed in molarity \((M)\), indicating moles of solute per liter of solution. Understanding solution concentration is crucial for preparing solutions and calculating reaction yields. In our case, to find the volume of silver nitrate needed, we used: - Known moles of silver nitrate \((0.004476 \text{ mol})\), with its solution concentration \((0.500 \text{ M})\). - We used the formula: \[ \text{Volume (L)} = \frac{\text{Moles of Solute}}{\text{Concentration (M)}} \] which gave us \(0.008952 \text{ L}\), or converting to milliliters, \(8.952 \text{ mL}\).This precise calculation ensures enough silver nitrate to fully react, forming the intended precipitate without excess.