A 2.559 9 pellet containing NaCl and KCl is dissolved in water. A silver nitrate solution is mixed with water containing the

Question

pellet, and a precipitate forms. The precipitate is separated and dried and found to have a mass of 5.112 9. What is the
percent composition of NaCl and KCl in the pellet? (Molar mass of NaCl = 58.44 g/mol, KCL = 74.55 g/mol, AgCl = 143.32 g/mol)
O 14.1% KCl and 85.9% NaCl
O 36.2% KCl and 63.8% NACI
o 50.0% KCl and 50.0% NaCl
63.8% KCl and 36.2% NaCl
85.9% KCl and 14.1% NACI

Answer 1

You will need two equations and they are solved simultaneously.
Let W = grams NaCl in the mixture
and Z = grams KCl in the mixture
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equation 1 is W + Z = 2.5599 g
Note: mm stands for molar mass
The second equation comes from the reaction with AgNO3.
AgNO3 + KCl = AgCl + KNO3 and
AgNO3 + NaCl = AgCl + NaNO3
egn 2 is grams AgCl from the KCl + grams AgCl from the NaCl = 5.1129
eqn 2 must be modified to be in terms of W and Z like this.
grams AgCl from KCl in terms of W: (W*mm AgCl/mmKCl)
grams AgCl from NaCl in terms of Z: (Z*mmAgCl/mmNaCl)
eqn 2 now is (W*mm AgCl/mmKCl) + (Z*mmAgCl/mmNaCl) = 5.1129
Solve equation 1 and equation 2 simultaneously to obtain W and Z.
Convert W and Z to percent of the sample like this.
% KCl = (W in grams/2.5599)*100 = ?
% NaCl = (Z in grams/2.5599)*100 = ?
Post your work if you get stuck.