Chemistry

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

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  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
    --------------------------------------------------
    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.

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    DrBob222
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