21 211 of 850 S ure Connection Substance Sought Substance Weighed Mn203 Ag2S CuC

Question

21 211 of 850 S ure Connection Substance Sought Substance Weighed Mn203 Ag2S CuCl2 Mn O Mn30 BaSO4 AgCl Pbl2 PROFESSOR's FAVORITE PROBLEM Contributed by Professor Thomas L. Isenhour, Old Dominion University to precipitate AgCI. After the precipitate is washed and dried, it weighs 21.62 g. What is the 76. A 10.00 g sample contains only NaCl and KCI. The sample is dissolved and AgNOis added weight percent of NaCl in the original sample? necommer 1. T. P. Hadjiioannou, G. D. Christian, C. E. Efstathiou, and D. Nikolelis, Problem Solving in Analytical Chemistry. Oxford: Pergamon, 1988

Explanation / Answer

Answer:

You have a mixture of sodium chloride and potassium chloride which you react, in aqueous solution, with silver nitrate, AgNO3.

The reaction will lead to the formation of silver chloride, a solid that will precipitate out of solution. The net ionic equation for this reaction describes what's going on

Ag+(aq) + Cl(aq) AgCl(s)

Notice that 1 mole of chloride ions is needed to produce 1 mole of silver chloride precipitate. This means that you can calculate the total number of moles of chloride that reacted by using the mass of the precipitate

(21.62g)(1 mole AgCl /143.32g) = 0.15085 moles AgCl In tables:Molar mass (AgCl) =143.32g/mol

Since you have a 1:1 mole ratio between silver chloride and the chloride ions, you'll have

Moles of Cl- = moles of AgCl = 0.15085 moles

This represents the total number of moles of chloride present in the sample. Since both sodium chloride and potassium chloride dissociate completely in aqueous solution, you can write

Moles of NaCl + moles of KCl = 0.15085 moles

If we call moles NaCl = X   and moles KCl = Y

X + Y = 0.15085   Eq. 1

One mole of NaCl produces 1 mole of Cl ions; the same is true for KCl.

The second equation will use the mass of the sample, which can be written like this

(moles NaCl)(MM NaCl) + (moles KCl).(MM KCl) =10.0g , where

MM NaCl, MM KCl are the molar masses of sodium chloride and potassium chloride, respectively.

If we call moles NaCl = X   and moles KCl = Y

X.(58.44) + Y(74.55) = 10.0g   Eq. 2

Now we solve the system using Eq 1 and Eq 2 :

X = 0.15085 + Y

58.44(0.15085 - Y) + 74.55 Y = 10.0

Y = 0.07351moles NaCl and X = 0.07734moles KCl

Now we can calculate mass of each one using molar mass:

mass of NaCl = (0.07351moles)(58.44g/mol) = 4.520g

mass of KCl = (0.07734moles)(74.55g/mol) = 5.480g

%NaCl= (4.520g/10.0g)x(100) = 45.2%

%KCl = (5.480g/10.0g)x(100) = 54.8%

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