Question

A 0.887 g sample of a mixture of NaCl and KCl is dissolved in water and the solution is treated with an excess of AgNO3 producing 1.822 g of AgCl. What is the percent by mass of each component in the mixture?

Answer 1

Two equations and two unknowns.
equation 1.
mass NaCl + mass KCl = 0.887 g
I would let y = mass NaCl and z = mass KCl.

equation 2.
AgCl from NaCl + AgCl from KCl = 1.822 g.
y g*(1 mol AgCl/1 mol NaCl) + z g*(1 mol AgCl/1 mol KCl) = 1.822.

Solve for y and z
You want percent of each; therefore,
(grams y/mass sample)*100 = %NaCl
(grams z/mass sample)*100 = %KCl.
[Note:Mass of the sample is 0.877 in the problem.][another note: where 1 mol AgCl/1 mol NaCl is used, substitute molar mass AgCl and molar mass NaCl. For 1 mol AgCl/1 mol KCl substitute molar mass AgCl and molar mass KCl.]

Answer 2

70.86

Answer 3

hi! in order to solve this you will need to incorporate a bit of algebra. first off, you have to produce the equations for each reactions.

NaCl + AgNO3 ---> AgCl + NaNO3
KCl + AgNO3 ---> AgCl + NaNO3

and let NaCl = x ; KCl = y

since the sample is equal to 0.887, and your sample contains only NaCl and KCl, then we have our first equation as:

x + y = 0.887 ----- (1)

and our precipitate is equal to 1.822. In order to complete the equation for the precipitate, we will relate NaCl to AgCl and KCl to AgCl by:

AgCl from NaCl:
x milligrams NaCl * (Mol wt AgCl/Mol wt NaCl)
x mg NaCl * (143.32 mg AgCl/58.44 mg NaCl) = 2.45x

and AgCl from KCl:
y milligrams KCl * (Mol wt AgCl/Mol wt KCl)
y mg NaCl * (143.32 mg AgCl/74.55 mg KCl) = 1.92y

with this, our second equation for the precipitate will be:
2.45x + 1.92y = 1.822 ---- (2)

Now, with our first and second equation, with two equations and 2 unknowns, input in your calculator and you will get an answer equal to:

x (NaCl) = 0.2245 mg
y (KCl) = 0.6625 mg

Now we can get the percent NaCl and KCl:

%NaCl = (0.2245/0.887)*100
%NaCl = 25.31% (ANSWER)

%KCl = (0.6625/0.887)*100
%KCl = 74.69% (ANSWER)

Hope this helped!! :)

Answer 4

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