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Darren needs to prepare 12 ounces of a 9% HCL solution. Find the amount of a 4% and the amount of 12% solution that he would need to mix to produce the needed solution.

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  • Lucy
    Lv 7
    1 decade ago
    Favorite Answer

    HINT: Write what you know in a table.

    Type …………. Amount …………. Percent …………. Total

    4% HCL …………. x …………. 0.04 …………. x * 0.04 = 0.04x

    12% HCL …………. y …………. 0.12 …………. y * 0.12 = 0.12y

    Total …………. x + y …………. 0.09 …………. (x + y) * 0.09 = 0.09(x + y)

    The first 2 rows of the last column always add up to the third row in that column.

    0.04x + 0.12y = 0.09(x + y)

    Given: Mixture is 12 ounces

    Means: x + y = 12

    Replace (x + y) with 12 and simplify.

    0.04x + 0.12y = 0.09(x + y)

    0.04x + 0.12y = 0.09(12)

    0.04x + 0.12y = 1.08

    You now have 2 equations.

    0.04x + 0.12y = 1.08

    x + y = 12

    Solve the 2nd one for y.

    x + y = 12

    y = 12 - x

    Replace y with 12 - x in the 1st one.

    0.04x + 0.12y = 1.08

    0.04x + 0.12(12 - x) = 1.08

    0.04x + 1.44 - 0.12x = 1.08

    0.04x - 0.12x = 1.08 - 1.44

    -0.08x = -0.36

    x = -0.36 / -0.08

    x = 4.5 = 4 1/2

    Substitute x with 4.5 in y = 12 - x to find y.

    y = 12 - x

    y = 12 - 4.5

    y = 7.5 = 7 1/2

    ANSWER: Darren needs 4 1/2 ounces of the 4% HCL solution and 7 1/2 ounces of the 12% HCL solution to make 12 ounces of a 9% HCL solution.

    Source(s): For more help with mixture word problems: http://www.purplemath.com/modules/mixture.htm
  • 1 decade ago

    The easiest way to do this is to calculate how much HCL and how much water you are working with.

    For example: in 12 ounces of a 9% solution, how much is HCL, and how much is water?

    Then figure out how much of the other two solutions you need to make it match that.

    There is likely a formula in your book, or a worked example, to show you how to combine two concentrations. It's a standard kind of problem that is often explained.

  • 1 decade ago

    balancing the amount of pure acid:

    pure HCL = (concentration as a decimal) * (volume)

    since there are 12 ounces total, if there are x ounces of 4% solution, there must be (12 - x) ounces of 12% solution)

    x ounces of 4% solution yields .04x ounces of HCL

    (12 - x) ounces of 12% solution yields .12(12 - x) ounces of HCL

    final solution has 12 ounces of 9% solution ==> .09(12) = 1.08 ounces of HCL

    balancing the inputs and output:

    .04x + .12(12 - x) = .09(12) = 1.08

    .04x + 1.44 - .12x = 1.08

    .36 = .08x

    x = .36 / .08 = 4.5 ounces

    therefore, 4.5 ounces of 4% solution (.18 ounces HCL)

    and 12 - 4.5 = 7.5 ounces of 12% solution (.9 ounces HCL)

    combine to give 12 ounces of 9% solution (1.08 ounces HCL)

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