chemistry

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.

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asked by mike
  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 = ?

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  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!! :)

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    posted by iza m.

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