Question

Using a recommended rate of 175 lbs N/acre (196 kg N/ha), develop a nitrogen fertilizer application for your 1.2m x 6.25m plot. How much more N fo you need to apply if 1 kg ammonium sulfate is applied to an area of 1.2m x 30m?

How much ammonium sulfate (21-0-0), complete fertilizer (14-14-14) or Sustane organic fertilizer (4-6-4) do you still need to apply to meet the remaining N requirements?

Answer 1

Ans. Given, Recommended supply rate of N = 196 kg N/ ha                ; [1 ha = 10⁴ m²]

                                                                        = 196 kg N/ 10000 m²

                                                                        = 0.0196 kg N/ m²

#1. Area of the plot = 1.2 m x 6.25 m = 7.5 m²

Recommended amount of N required for plot =

= Recommended N supply rate x Area of plot

                                                = (0.0196 kg N/ m²) x 7.5 m²

                                                = 0.147 kg

Ammonium sulfate, (NH₄)₂SO₄ has molar mass of 132.14 g/mol.

1 mol ammonium sulfate (=132.14 g) has 1 mol (= 14.00 g) N.

            Or,       132.14 g Ammonium sulfate is equivalent to 14 g N

            Or,       1.00 g             -                       -           -           -    (14 / 132.14) g N

            Or,       1000 g            -           -           -           -           - (14 / 132.14) x 1000 g N  

                                                                                                = 105.95 g

                                                                                                = 0.10595 kg                       

Therefore, 1 kg ammonium sulfate (AS) is equivalent to 0.106 kg N.

Now,

            Additional amount of N needed = Recommended N amount – N supplied from AS

                                                                        = 0.147 kg – 0.106 kg

                                                                        = 0.041 kg

#2.A. The 21-0-0 fertilizer means that there is 21% N and 0% each of P and K.

Let the amount of its required be X kg.

Now,

            Nitrogen content in fertilizer = 21 % of X kg = 0.21X kg

This 0.21X kg shall be equal to the amount of N additionally required.

            So,

                        0.21X kg = 0.041 kg

                        Or, X = 0.041 / 0.21 = 0.195 kg

Therefore, required amount of 21-0-0 fertilizer = 0.195 kg

#2.B. Let the amount of 14-14-14 fertilizer be X kg.

Now,

            Nitrogen content in fertilizer = 14 % of X kg = 0.14X kg

This 0.14X kg shall be equal to the amount of N additionally required.

            So,

                        0.14X kg = 0.041 kg

                        Or, X = 0.041 / 0.14 = 0.293 kg

Therefore, required amount of 14-14-14 fertilizer = 0.293 kg

#2.C. Let the amount of 4-6-4 fertilizer be X kg.

Now,

            Nitrogen content in fertilizer = 4 % of X kg = 0.04X kg

This 0.04X kg shall be equal to the amount of N additionally required.

            So,

                        0.04X kg = 0.041 kg

                        Or, X = 0.041 / 0.04 = 1.025 kg

Therefore, required amount of 4-6-4 fertilizer = 1.025 kg