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

1. 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?  2.

3. 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.  4. 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 didn´t see that the 18.2 mL are per 1g of cyanide..

Bye!  5. Originally Posted by M_Gabriela 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 didn´t 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?  6. Because I confused!! I didn´t 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 don´t 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 I´m not sure.

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

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

bye  7. Originally Posted by M_Gabriela Because I confused!! I didn´t 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 don´t 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 I´m not sure.

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

Well, now I´m 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.  8. Because I was considering mg instead of g. But % m/V is gram of solute in 100 mL of solution. Uff.. serious mystake. I don´t 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!  Bookmarks
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