Gen. Chemistry

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?

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  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.]

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  2. 70.86

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  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!! :)

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