4. How much potassium nitrate (KNO,) should be dissolved in a 100-gallon stock (

Question

4. How much potassium nitrate (KNO,) should be dissolved in a 100-gallon stock (installed with a 1:200 ratio injector) to provide 14 meq/L NO, in the final fertilizer solution? Answer: Amount of KNO) needed per 100 gallons- lb- kg Calculations If you have 12 meqyL. NO, in your fertilizer solution, what would be the concentration of and nitrate (NO,) in parts per million? Answers: 1) Concentration of N 2) Concentration of NO Ppm NO Molybdenum is required for the normal growth and development of poinsettia. How much molybdenum trioxide (MoO,) should be used for a 50-gallon stock tank installed with a 1:200 ratio injector to provide 0.05 ppm Mo in the final fertilizer solution? The molecular weight of molybdate (MoO,) is 144. The atomic weights of Mo and O are 96 and 16, respectively 6. -g/50-gallon stock tank. Answer: Amount of molybdenum trioxide (MoO) Calculations:

Explanation / Answer

4. 235.895 lb = 107 kg

5. Mass of NO3- in = 744 ppm, Mass of N = 168 ppm

6. 2.839 g MoO3 / 50 gallon tank

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1 Gallon = 3.78541 L

1 lb = 0.453592 kg

M.W. KNO3 = 101.10 g/mol

M.W. NO3- = 62.00 g/mol

1 eq / L of NO3- = 62.00 g/L

1 meq / L of NO3- = 62 mg/L

14 meq/L of NO3- = 14 x 62 = 868 mg/L

Since one KNO3 give one NO3- ions

Amount of KNO3 required to provide 14 meq/L NO3- = 14 x 101.10 = 1,415.4 mg / L

so amount of KNO3 required for 100 gallons = 100 x 3.78541 x 1,415.4 = 535,787 mg = 0.535 kg = 1.181 lb

Since it is connected with 1:200 injector, the amount of KNO3 required make the stock = 200 x 0.535 = 107 kg = 235.895 lb

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Amount of NO3- = 12 meq/L

M.W. of NO3- = 62.00 g/mol

Atomic Wt of N = 14.00 g/mol

Mass of NO3- in 12 meq/L solution = 12 x 62.00 = 744 mg/L = 744 ppm ( 1 mg/L = 1 ppm)

Mass of N in 12 meq/L solution = 12 x 14 = 168 mg/L = 168 ppm

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M.W. MoO3 = 144 g/mol

At. Wt Mo = 96 g/mol

At wt. O = 16 g/mol

Concentration of the Mo in 50 gallon = 0.05 ppm = 0.05 mg/L

Equivalence of Mo = 0.05 / 96 = 0.0005208 meq/L

Weight of MoO3 in 0.0005208 meq/L = 0.0005208 x 144 = 0.0749952 mg /L

Since the injector dilution is 1:200, the stock concentration = 200 x 0.075 mg/L = 15.0 mg /L

Stock tank volume = 50 gallon, so the weight of MoO3 in 50 gallon tank = 50 x 3.78541 L x 15.0 mg/L = 2839 mg = 2.839 g / 50 gallon tank

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