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Thread: can people just check a calculation for me because it seems a bit off

  1. #1 can people just check a calculation for me because it seems a bit off 
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    Ok,
    so I have a solution of 4% cyanide in 5000 liters of water.
    Using NaOCl to neutralize the cyanide 18.2 ml is required per gram of cyanide. how much NaOCl is required to destroy the cyanide?

    now I have taken % to be parts per hundred and converted that into ppm (mg/kg) as the majority of the solution is water and the specific gravity of water is 1 it is 4x10000 to get 40000 mg/kg and then multiplied this by 5000 to get mg in 5000 liters. this has then been divided by 1000 to get grams.

    then I have done this answer multiplied by 18.2.
    my end answer is 3640000 ml or 3640 liters.

    have I done this correctly? the value seems awfully high to me. if not where have I gone wrong?


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  3. #2  
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    Seems ok to me.
    Consider 100 ml of 4% solution. That would contain 4g of cyanide, requiring 72.8 ml of neutralising solution.
    So the volume of neutralising solution is just a little less than the volume of cyanide solution.


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  4. #3  
    Forum Sophomore M_Gabriela's Avatar
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    You have 0.200 g (or 200 mg) of cyanide in 5000 mL
    18.2 mL of NaOCl neutralize 1 g

    So you would need
    1 g ---- 18.2 mL
    0.2 g-----x=3.64 mL

    You didnīt see that the 18.2 mL are per 1g of cyanide..

    Bye!
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  5. #4  
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    Quote Originally Posted by M_Gabriela View Post
    You have 0.200 g (or 200 mg) of cyanide in 5000 mL
    18.2 mL of NaOCl neutralize 1 g

    So you would need
    1 g ---- 18.2 mL
    0.2 g-----x=3.64 mL

    You didnīt see that the 18.2 mL are per 1g of cyanide..

    Bye!
    Could you please explain how you arrive at 200 mg in 5000 ml? And why you have done a calculation for 5000 ml, when the question was about 5000 litres?
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  6. #5  
    Forum Sophomore M_Gabriela's Avatar
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    Because I confused!! I didnīt see 5000 L, haha, I saw mL.

    So, if you take

    0,004 g cyanide ---- 0,1 L (100 mL)
    200 g cyanide ----- 5000 L

    Though I have to admit that I donīt know if it is correct to use this because it says 5000 L of water (solvent) not solution. If it is done properly, I would need the density of the cyanide solution. But because the percentage is low... may be is not significant... Though Iīm not sure.


    Then
    1 g ---- 18,72 mL
    200 g ---- x= 3640 mL

    Well, now Iīm not sure. How did you get those 72.8 mL of NaOCl?

    bye
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  7. #6  
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    Quote Originally Posted by M_Gabriela View Post
    Because I confused!! I didnīt see 5000 L, haha, I saw mL.

    So, if you take

    0,004 g cyanide ---- 0,1 L (100 mL)
    200 g cyanide ----- 5000 L

    Though I have to admit that I donīt know if it is correct to use this because it says 5000 L of water (solvent) not solution. If it is done properly, I would need the density of the cyanide solution. But because the percentage is low... may be is not significant... Though Iīm not sure.


    Then
    1 g ---- 18,72 mL
    200 g ---- x= 3640 mL

    Well, now Iīm not sure. How did you get those 72.8 mL of NaOCl?

    bye
    I got the 72.8 ml from 4g x 18.2 ml/g

    How do you get .004 g of cyanide in 100 ml? Why not 4 g?

    I think we can ignore the density of the solution.
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  8. #7  
    Forum Sophomore M_Gabriela's Avatar
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    Because I was considering mg instead of g. But % m/V is gram of solute in 100 mL of solution. Uff.. serious mystake. I donīt know what happened there..,

    Thanks.

    So 4 g in 100 mL.... that would give 200 Kg in 5000 L

    So if 1 g is neutralized with 18.2 mL, then 200 kg needs 3640 L.

    So, James did it correctly! Bravo!
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