A mixture of NaCl and KCl weighed 5.4892 g . The sample was dissolved in H2O and reacted with excess AgNO3 to form AgCl. This AgCl weighed 12.7052 g. Wht was th

mass of NaCl and KCl : 5,4892 gram
mass of AgCl : 12,7052 gram
Concentration of NaCl in the sample : ... %

Answered

step 1
Make a reaction first. There are 2 type of the reaction.

1. A mixture of NaCl and KCl with H2O. H2O will decompose of the mixture to be NaCl + KCl

Reaction
NACl. KCl → NaCl + KCl

2. NaCl and KCl added AgNO3 leads to precipitate to the formation of AgCl

Reaction
NaCl + AgNO3 → AgCl + NaNO3

KCl + AgNO3 → AgCl + NaNO3

Step 2
Determine mole of AgCl

mole : mass / Mr

mole of AgCl : 12,7052 g / 143,5 g/mole
AgCl : 0,0885 moles

We has found out mole of AgCl. So, we can determine mole of Cl- in AgCl

Ag+ + Cl- → AgCl

mole of Cl- : (c. Cl-/c. AgCl) x moles of AgCl
Cl- : (1/1) x 0,0885 moles
: 0,0885 moles

step 3
Make a equation. There are 2 equation in the question. We know that NaCl and KCl dissociate completely in aqueous solution, so we can write the equation :
a : NaCl
b : KCl

The first equation
a + b : 0,0885

The second equation
If we want to make the second equation. We know the mixture contains NaCl and KCl. We have to convert moles of NaCl and KCl to gram

NaCl : a mole
Mr NaCl : 58,5 g / mole

KCl : b mole
Mr KCl : 74,5 g/mole

g NaCl : a mole x 58,5 g/mole
: 58,5a gram

g KCl : b mole x 74,5 g/mole
: 74,5b gram

So, the second equation is

58,5a + 74,5b : 5,4892

step 4
Determine value of a and b. By using the substitution method.

a + b : 0,0885
a : 0,0885 - b

58,5a + 74,5b : 5,4892
58,5(0,0885 - b) + 74,5b : 5,4892
5,1772 - 58,5b + 74,5b : 5,4892
16b : 5,4892 - 5,1772
b : 0,312 / 16
b : 0,0195

58,5a + 74,5b : 5,4892
58,5a + 74,5(0,0195) : 5,4892
58,5a + 1,4528 : 5,4892
a : 4,0364 / 58,5
a : 0,0690

step 5
Determine mass of NaCl. Substitution of value a so we can find out mass of NaCl

g NaCl : a mole x 58,5 g/mole
g NaCl : 0,0690 mole x 58,5 mole
: 4,0365 gram

step 6
Determine concentration of NaCl in the mixture.

% NaCl : (mass of NaCl / mass sample) x 100%
% NaCl : (4,0365 g / 5,4892 g) x 100%
: 73,53% (b/b)

So, the concentration of NaCl in the mixture is 73,53% (b/b)

Source :
socratic.org

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