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Math Help - Chemical solution question.

  1. #1
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    Chemical solution question.

    A chemical manufacturer has an order for 500 litres of a 25% acid solution (i.e. 25% by volume is acid). Solutions of 30% and 18% are available in stock.
    How much acid is required to produce 500 litres of 25% acid solution?
    The manufacturer wishes to make up the 500 litres from a mixture of 30% and 18% solutions.
    Let x denote the amount of 30% solution required.
    Let y denote the amount of 18% solution required.
    Use simultaneous equations in x and y to determine the amount of each solution.
    I'm pretty sure once I find what the simultaneous equations are, I will be able to solve them.
    Thanks.
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  2. #2
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    Quote Originally Posted by grooverandshaker View Post
    A chemical manufacturer has an order for 500 litres of a 25% acid solution (i.e. 25% by volume is acid). Solutions of 30% and 18% are available in stock.
    How much acid is required to produce 500 litres of 25% acid solution?
    The manufacturer wishes to make up the 500 litres from a mixture of 30% and 18% solutions.
    Let x denote the amount of 30% solution required.
    Let y denote the amount of 18% solution required.
    Use simultaneous equations in x and y to determine the amount of each solution.
    I'm pretty sure once I find what the simultaneous equations are, I will be able to solve them.
    Thanks.
    1. The 1st equation refers to the volumes of the liquids:

    x+y=500

    2. The 2nd equation refers to the volume of the acid:

    0.3 \cdot x + 0.18 \cdot y  = 0.25 \cdot 500

    3. Solve for x and y.
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  3. #3
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    mixture problem

    an easier to way to solve this

    draw a concentration line

    18 25 30
    lean mix rich
    7 5

    5 parts lean + 7 parts rich = 12 parts mix

    208.3 liter + 291.6 liter = 499.9liters mix



    bjh
    Last edited by bjhopper; January 30th 2011 at 04:15 PM. Reason: correct decimal
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  4. #4
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    Thank you for all of your help! =)
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