Advance Study Assignment A student obtained a vial of powder consisting of a mix

Question

Advance Study Assignment A student obtained a vial of powder consisting of a mixture of iron (III) chloride hexahydrate (FeCI_2*6H_2O) and silver nitrate (AgNO_3) in unknown proportions. He weighed the vial with the powder in it and found the mass to be 20.560 g. He then poured the powder into a beaker and found the mass of the empty vial to be 15.723 g. Water was added (500 ml) to powder in the beaker, and the contents were mixed and heated for 15 minutes. Once the reaction was complete, a silver chloride precipitate was left in the water. The student then obtained a piece of filter paper with a mass of 1.043 g, put it in a funnel, and poured the contents of the beaker through it. Once the filter paper an precipitate were dried in an oven, the mass of both together was found to be 2.778 g. What was the mass of the unknown powder in the vial? Write balanced chemical equation for the reaction that occurs between iron (III) chloride hexahydrate and silver nitrate (include phases). What is the molar mass of the following: What was the mass of the silver chloride produced? How many moles of silver chloride were produced? Use the chemical equation from to determine the mass of iron(III) chloride hexahydrate needed to produce the observed amount of silver chloride, Use the chemical equation from to determine the mass of silver nitrate needed to produce the observed amount of silver chloride What was the total mass of reacted salts (use the quantities from (6) and (7)? How many grams of excess reactant were present in the initial sample?

Explanation / Answer

Answer – Given, mass empty vial = 15.723 g ,mass of unknown powder+ vial =20.560 g

Mass of filter paper = 1.043 g ,

mass of filter paper + precipitate of silver chloride = 2.778 g

1)Mass of the unknown powder in the vial = mass of unknown powder+ vial - mass empty vial

                                                                     = 20.560 g -15.723 g

                                                                     = 4.837 g

2) Reaction between iron (III) chloride and silver nitrate

FeCl3.6H2O + 3 AgNO3 -----> 3 AgCl(s) + Fe(NO3)3 + 6H2O

3) Molar mass of each reaction as follow –

FeCl3.6H2O = 270.30 g/mol

AgNO3 = 169.87 g/mol

AgCl = 143.32 g/mol

Fe(NO3)3 = 241.86 g/mol

4) Mass of silver chloride = mass of filter+ precipitate – mass of filter paper

                                          = 2.778 – 1.043

                                          = 1.735 g

5) moles of silver chloride = 1.735 g / 143.32 g.mol-1

                                       = 0.0121 moles

6) From the balanced reaction –

3 moles of AgCl = 1 moles of FeCl3.6H2O

So, 0.0121 moles AgCl = ?

= 0.00404 moles

So, mass of FeCl3.6H2O = 0.00404 moles * 270.30 g/mol

                                       = 1.09 g

7) From the balanced reaction –

3 moles of AgCl = 3 moles of AgNO3

So, 0.0121 moles AgCl = ?

= 0.0121 moles

So, mass of AgNO3= 0.01251 moles * 169.87 g/mol

                                = 2.06 g

8) mass of reactant reacts = 1.09 g + 2.06 g = 3.147 g

9) mass of excess of reactants = 4.837 g – 3.147 g

                                                = 1.690 g

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