91Ó°ÊÓ

Water soluble fertilisers are nutrients that are completely soluble in water and are administered by Foliar Spray & Fertilizers procedures, allowing nutrients to be rapidly absorbed and utilised in the plant system without evaporation, leaching, or runoff.

(a)

Compute the molar masses of each combination, then divided the atomic mass of$\mathrm{P}$by the molar mass of the compound to get the mass percent of $\mathrm{P}$.

Let’s calculate the molar masses of ${\left({\text{NH}}_4\right)}_2{\text{HPO}}_4$

$$\begin{gathered} M\left({\left(N{H}_4\right)}_{2}HP{O}_4\right)=2A\left(N\right)+9A\left(H\right)+A\left(P\right)+4A\left(O\right) \\ M\left({\left(NH\right)}_{2}HP{O}_4\right)=132.06gmo{l}^{-1} \\ w\left(P,{\left(N{H}_4\right)}_{2}HP{O}_4\right)=\frac{A\left(P\right)}{M\left({\left(NH\right)}_{2}HHO{O}_4\right)} \\ w\left(P{\left(N{H}_4\right)}_{2}HPO\right)=\frac{30.974}{132.06}×100\% \\ w\left(P,{\left(N{H}_4\right)}_{2}HP{O}_4\right)=23.45\% \end{gathered}$$

Then calculate the molar masses of ${\text{NH}}_4\left({\text{H}}_2{\text{PO}}_4\right)$

$$\begin{gathered} M\left(N{H}_4\left(H_{2}P{O}_4\right)\right)=A\left(N\right)+6A\left(H\right)+A\left(P\right)+4A\left(O\right) \\ M\left(N{H}_4\left(H_{2}P{O}_4\right)\right)=115.03gmo{l}^{-1} \\ w\left(P,N{H}_4\left(H_{2}P{O}_4\right)\right)=\frac{A\left(P\right)}{M\left(N{H}_4\left(H_{2}P{O}_4\right)\right)} \\ w\left(P,N_4\left(H_{2}P{O}_4\right)\right)=\frac{30.974}{115.03}·100\text{\%} \\ w\left(P,N_4\left(H_2{P}_4\right)\right)=26.93\text{\%} \end{gathered}$$

From the calculation, we observed that ${\text{NH}}_4\left({\text{H}}_2{\text{PO}}_4\right)$has higher mass percentage than $\mathrm{P}$.

b)

The overuse of soluble phosphate fertilisers has become a major source of phosphorus cycle imbalance.

The fertiliser with the lowest mass percentage of $\mathrm{P}$ is preferred to reduce that undesirable effect.