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Re: Chemistry Help





Hi Steve,

Use this calculation:

Molecular Wt of KNO3 = 101.1 g/mol
Molecular Wt of NO3+ = 62.0 g/mol

(1) 36.6 g (KNO3) * 1 mol/101.1 g = 0.362 moles of KNO3 ==> 0.362 moles of NO3+

    0.362 mol (NO3+) * 62.0 g/1 mol = 22.44 g (NO3+) ==> 22440 mg (NO3+)

    Therefore, [NO3+] = 22440 mg/200 mL = 112.2 mg/mL ==> 112200 ppm

(2) If take 1.0 mL and add to 100 L, then:

    112.2 mg/mL * 1.0 mL = 112.2 mg

    Therefore, [NO3+] = 112.2 mg/100 L = 1.122 mg/L ==> 1.122 ppm

For a nitate concentration of 5 ppm, use 5 mL of your stock solution:

    112.2 mg/mL * 5.0 mL = 561.0 mg

    Therefore, [NO3+] = 561.0 mg/100 L = 5.61 mg/L ==> 5.61 ppm

Hope this helps.

Peter
Ottawa, Canada

:: I am trying to figure out how much KNO3 to add to my tank so I can get about
:: 5ppm of NO3 but  chemistry is not that great.  I gonna write what I got so
:: far, and if you all can tell me if I'm on the right track.
::
:: 3 tablespoon KNO3 ~ 36.6g
:: 200 ml water
::
:: Concentration of KNO3 in 200ml water - 36600 / 0.2 = 180000ppm
::
:: NO3 is 61% off KNO3.. so
::
:: 180000 / 0.61 = 109800ppm of NO3 ....... right?
::
:: so if I take 1ml off this solution and add it to my 100 liter tank,  What
:: will I get?
::
:: I can't seem to wrap my mind around the math on this one.
::
:: is this close?
::
:: (109800/1000) / 100l = 1.098 ppm
::
:: Thanks
:: Steve (whos head is big as a pumpkin right now)