Question

What is the actual amount of phosphorous in a fertilizer with a npk ratio of 15-15-20?

Answer 1

The actual amount of phosphorus in fertilizer with NPK ratio of 15-15-20 is $\boxed{{\text{65}}{\text{.51 g}}}$.

Further explanation:

NPK ratio refers to the ratio of nitrogen, phosphorus, and potassium in any fertilizer. This ratio represents the percentage of corresponding nutrient present in the fertilizer. It actually determines the amount of ${{\text{P}}_2}{{\text{O}}_5}$.

The given fertilizer has the NPK ratio of 15-15-20. This implies the percentage of nitrogen in the fertilizer is 15 %, that of phosphorus is 15 % and that of potassium is 20 %.

Consider the mass of ${{\text{P}}_2}{{\text{O}}_5}$  to be 1 kg.

The mass of P is calculated as follows:

${\text{Mass of P}}=\left({\dfrac{{{\text{\% of P}}}}{{{\text{100 \% }}}}}\right)\left({{\text{Mass of }}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}}\right)$         ...... (1)

Substitute 15 % for % of P and 1 kg for the mass of ${{\text{P}}_2}{{\text{O}}_5}$ in equation (1).

$\begin{aligned}{\text{Mass of P}}&=\left({\frac{{{\text{15 \%}}}}{{{\text{100 \% }}}}}\right)\left( {{\text{1 kg}}} \right)\\&=0.1{\text{5 kg}}\end{aligned}$

The formula to calculate the amount of ${{\text{P}}_2}{{\text{O}}_5}$  is as follows:

${\text{Amount of }}{{\text{P}}_2}{{\text{O}}_5}=\dfrac{{{\text{Given mass of }}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}}}{{{\text{Molar mass of }}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}}}$       ...... (2)

The mass of ${{\text{P}}_2}{{\text{O}}_5}$ is to be converted into g. The conversion factor for this is,

${\text{1 kg}}={\text{1}}{{\text{0}}^{\text{3}}}{\text{ g}}$

So the mass of ${{\text{P}}_2}{{\text{O}}_5}$ is calculated as follows:

$\begin{aligned}{\text{Mass of }}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}&=\left( {{\text{0}}{\text{.15 kg}}}\right)\left({\frac{{{\text{1}}{{\text{0}}^{\text{3}}}{\text{ g}}}}{{{\text{1 kg}}}}}\right)\\&={\text{150 g}}\end{aligned}$

The given mass of ${{\text{P}}_2}{{\text{O}}_5}$ is 150 g.

The molar mass of ${{\text{P}}_2}{{\text{O}}_5}$ is 141.9445 g/mol.

Substitute these values in equation (2).

$\begin{aligned}{\text{Amount of }}{{\text{P}}_2}{{\text{O}}_5}&=\left( {{\text{150 g}}} \right)\left({\frac{{{\text{1 mol}}}}{{{\text{141}}{\text{.9445 g}}}}}\right)\\&={\text{1}}{\text{.05675 mol}}\end{aligned}$

The chemical formula ${{\text{P}}_2}{{\text{O}}_5}$ indicates that it consists of two moles of P. So mass of P is calculated as follows:

${\text{Mass of P}}=\left({{\text{Amount of }}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}} \right)\left[{2\left( {{\text{Molar mass of P}}}\right)}\right]$            ...... (3)

The amount of ${{\text{P}}_2}{{\text{O}}_5}$ is 1.05675 mol.

The molar mass of P is 31 g/mol.

Substitute these values in equation (3).

$\begin{aligned}{\text{Mass of P}}&=\left({{\text{1}}{\text{.05675 mol}}}\right)\left[{2\left({\frac{{{\text{31 g}}}}{{{\text{1 mol}}}}}\right)}\right]\\&={\text{65}}{\text{.51 g}}\end{aligned}$

So the actual amount of phosphorus in the fertilizer is 65.51 g.

Learn more:

1. What happens to the reducing agent during redox reaction?: brainly.com/question/2890416

2. Identification of element which has electron configuration  :brainly.com/question/9616334

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Environmental chemistry

Keywords: phosphorus, P2O5, fertilizer, 100 kg, actual amount, NPK, 15-15-20, 15 %, 20 %, 15 %.

Answer 2

NPK ratio system is a conventional shorthand for the ratio of the nitrogen N), phosphorus (P) and potassium (K) in a fertilizer. It actually shows amount of P₂O₅.
If we use 100 kg of P₂O₅:
m(P₂O₅) = 0,15 · 100 kg.
m(P₂O₅) = 15 kg.
m(P₂O₅) : M(P₂O₅) = m(P) : 2M(P).
15 kg : 142 = m(P) : 62.
m(P) = 6,55 kg.
ω(P) = 6,55 kg ÷ 100 kg · 100% = 6,55%.