Question

1. An 887.0 mg sample of a mixture containing only NaCl and KCl is dissolved in water, and excess AgNO3 is added to yield 1.913 g of AgCl. Compute the mass-% of each compound in the mixture.

Answer 1

Set up two equation. They are
(1) g NaCl + g KCl = 0.8870g
(2) g AgCl from NaCl + g AgCl from KCl = 1.913g

If you let X = g NaCl
and let Y = g KCl, then equation 1 becomes X+Y = 0.8870 and equation 2 becomes
X(MMAgCl/MMNaCl) + Y(MMAgCl/MMKCl) = 1.913 where MM stands for molar mass.

Solve the two equation simultaneously for X and Y, then
%NaCl = (X/0.8870)*100 = ?
%KCl = (Y/0.8870)*100 = ?

Answer 2

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 1.913. 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 = 1.913 ---- (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) = 0.3962 mg
y (KCl) = 0.4908 mg

Now we can get the percent NaCl and KCl:

%NaCl = (0.3962/0.887)*100
%NaCl = 44.67% (ANSWER)

%KCl = (0.4908/0.887)*100
%KCl = 55.33 (ANSWER)

Hope this helped!! :)