Question

A Crop of Maize needs 5 kg of P₂O₅ /ton of grain. If the yield goal is 12 ton/ha and the soil has a concentration of 10 ppm of P₂O₅, how many kg of Diammonium phosphate (DAP fertilizer with 50% P) must be added in the basal fertilizer to meet the crop demand? Assume we apply all 100% DAP at one time

Answer 1

Task #80857

A Crop of Maize needs 5 kg of P₂O₅ /ton of grain. If the yield goal is 12 ton/ha and the soil has a concentration of 10 ppm of P₂O₅, how many kg of Diammonium phosphate (DAP fertilizer with 50% P) must be added in the basal fertilizer to meet the crop demand? Assume we apply all 100% DAP at one time.

Solution.

Firstly, we need to find how many m⁺(P₂O₅) we need for the harvest.

$$\begin{array}{l} P_2O_5 + 3H_2O + 4NH_3 = 2(NH_4)_2HPO_4 \\ m^+(P_2O_5) = 5 \text{ kg/ton} * 12 \text{ ton} = 60 \text{ kg} \end{array}$$

Secondly, since 10 ppm phosphorus is present in the soil, we must dissolve the resulting mass by the concentration of phosphorus in the soil. (10 ppm = 0.005%)

$$\begin{array}{l} m_1(P_2O_5) = 60 / 0.001 = 60000 \text{ kg} \\ n_1(P_2O_5) = 422535 \text{ mole} \\ n((NH_4)_2HPO_4) = 845070 \text{ mole} \\ m((NH_4)_2HPO_4) = 97183.05 \text{ kg} \end{array}$$

Answer:

$$m((NH_4)_2HPO_4) = 97183.05 \text{ kg}$$

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