Help with mixing equation

Discussion in 'Pesticide & Herbicide Application' started by Slyder777, Sep 16, 2007.

  1. Slyder777

    Slyder777 LawnSite Member
    from Texas
    Posts: 22

    OK, sorry for the noob post, but I am missing something here and need a pro to tell me what it is.

    I am studying for my license exam and this is an exact question on the mixing part of the study guide.

    You need to apply a pesticide premise spray for livestock pests to an area of 70,000 sq. ft. The pesticide is a 57% EC (5 lb/gal.) formulation. The label recommended rate is 1 ounce per 3,000 sq. ft. and requires an application of a 2 percent spray mixture in an oil carrier (oil weight = 7.05 lb/gal).

    A. How many ounces of pesticide concentrate do you need to treat the premises?

    B. How many gallons of oil carrier do you need to prepare the 2 percent mixture?

    Question A is an obvious and easy answer. Question B is throwing me for a loop. I have the answer, and still can't get how they got to it. I have had no other problems with any of the mix eqautions to this point.

    Also, the first two answers they provide in the book were wrong. They had to post a correction with two new answers. I am starting to wander is their answer for B is still wrong.

    Again, I know the answer, but need to know how to get there. THanks for any help.

    Shawn
     
  2. americanlawn

    americanlawn LawnSite Fanatic
    from midwest
    Posts: 5,860

    Seems they are trying to throw you for a loop (trick question). Multiply the answer of question A times .02

    That's my best guess.
     
  3. mkroher

    mkroher LawnSite Senior Member
    Posts: 539

    hi. it says the "label recommends a rate of 1oz per 1000sqft". I think that means solution, so the info about the 57% stuff is non important.

    A). 1oz/3000sqft * 70,000sqft = 23.3oz sol

    B) 23.3oz sol * 1 gal/sol / 128oz * 100gal oil / 2 gal sol = 9.10 gal oil ??

    That's my guess.
     
  4. RigglePLC

    RigglePLC LawnSite Fanatic
    Posts: 12,353

    I will try. I agree, 23.3 ounces concentrate needed.

    For part B. 23.3 ounces will be 2 percent of your new oil solution.

    Divide 23.3 by .02

    Result 116.5 ounces. The total amount to be applied.

    Divide by 128. You get .91 gallon of oil carrier required.
     
  5. Leaf Jockey

    Leaf Jockey LawnSite Senior Member
    Posts: 361

    My turn.
    1 ounce concentrate added to 49 ounces of carrier is a 2% solution.
    49 times 23.3 = 1141.7 ounces
    divide by 128 = 8.91

    Scott
     
  6. Slyder777

    Slyder777 LawnSite Member
    from Texas
    Posts: 22

    WEll, at least I know I'm not the only one who had trouble with this problem. I am really wondering if the question is worded wrong. Leaf Jocky is the closest so far. I will wait for a few more replies and then post what the book says is the correct answer for B. 23.3 oz is correct for A.

    I have disected this thing from front to back, inside and out and can't seem to figure out how they got the answer they did. Of course the reason that worries me is I do not want to run into a problem like this in the course of a day and not know how to resolve it.

    Thanks again for the replies. Let's hope someone on here can figure this thing out.

    What is really frustrating is most other equations and problems were explained and talked about. Nothing in all the study material talks about a problem like this.
     
  7. RigglePLC

    RigglePLC LawnSite Fanatic
    Posts: 12,353

    Whoops! My mistake.

    23.3 divided by .02 is 1165 ounces

    Divide by 128 to get 9.1 gallons.
     
  8. Leaf Jockey

    Leaf Jockey LawnSite Senior Member
    Posts: 361

    I think you need to subtract the 23.3 ounces from the 1165 so your answer reflects just the amount of carrier. Thats how I figure it anyway.

    Scott
     
  9. Slyder777

    Slyder777 LawnSite Member
    from Texas
    Posts: 22

    OK, finally figured it out.

    Here is the equation given to use for EC mixing using percentages.

    gals of EC needed = (gals of spray wanted) x (% of ai wanted) x (lbs/gal of carrier) divided by lbs ai per gallon x 100

    So here is what the equation should look like using the information we already know.

    .18 = X(Gals of spray wanted) x 2 x 7.05 divided by 5 x 100

    .18 = 14.10x divided by 500

    .18 = .0282X

    You then use basic algebra and divide .18 by .0282 to find X which is 6.38. 6.38 is the answer given in the book.

    So you need 6.38 gallons of oil to carry the 23.3 ounces of ai to treat the 70,000 sqaure feet. Whew.

    Knew it was there somewhere, but dang, didn't think it would be that hard.

    BTW .18 was achieved by dividing the total ounces 23.3 Ounces by 128 to determine the number of gallons of ai needed.

    Thanks for the help guys. Now we know how to get from poiont A to Point B via Point R O X D U V C. :dizzy: Have a good night.

    Shawn
     
  10. RigglePLC

    RigglePLC LawnSite Fanatic
    Posts: 12,353

    Mkrohr is right. Ok, but problem didn't mention they wanted 2 percent by weight.

    Convert 23 ounces to pounds by dividing by 128, getting .1796

    Multiply by 5 to get the weight of pesticide, .8984 lbs

    Divide by the percent .02 to get pounds of oil, 44.9

    Divide by 7.02 to get gallons of oil, 6.399
     

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