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Chapter 4: Problem 108

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

Expert verified
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

01

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.
02

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
03

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
04

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
05

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%.
06

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
07

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.

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