Question

Consider a 1.24 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 0.419 g. Calculate the mass percent of magnesium chloride in the mixture. % MgCl2 Determine the minimum volume of silver nitrate that must have been added to ensure complete formation of the precipitate. mL AgNO3

Answer 1

Answer:

Explanation:

Step 1: Balance the reaction

MgCl2 (aq) + 2 AgNO3 (aq)  → 2AgCl (s) + Mg(NO3)2 (aq)

For 1 mole MgCl2 we have 1 mole Mg(NO3)2, but 2 moles AgNO3 and 2 moles AgCl

Step 2 : Calculating moles

⇒Mass of magnesium nitrate (Mg(NO3)2 )  and magnesium chloride (MgCl2) = 1.24g

⇒Molarity of the silver nitrate (AgNO3) = 0.5M

⇒The mass of the white precipitate formed is 0.419 g

mole AgCl = mass AgCl / Molar mass AgCl

mole AgCl = 0.419g / 143.32g/mole = 0.002935 mole

Step3 : Calculating mass

For 2 mole AgCl we have 1 mole MgCl2, so to find the mole of MgCl2 we have to divide by 2

mole MgCl2 = 0.002935 mole/2 = 0.0014618 mole

mass MgCl2 = 0.0014618 mole * 95.211g/mole =0.13918 g

Mixture has a mass of 1.24g

Mgcl2 has a mass of 0.13918g

(0.13918 / 1.24) x 100% = 11.224 %

⇒11.224% of the mixture is MgCL2

b) 0.002935 mole of AgCl gives 0.002935 mole of AgNO3

Molarity = moles / volume  or moles = molarity * volume

⇒ 0.002935 moles = 0.500 M * volume

volume = 0.002935 moles / 0.5 M = 0.00587 L = 5.87 mL

The minimum volume of silver nitrate that must have been added = 5.87 mL