# For phosphoric fertilizer we calculate the part of the weigh of P2O5 and potassium weight of K2O. We know that 1 ha uses 154kg of this fertilizer (N + P2O5 + K2O in ratio 3:1:3). Let's say we use ammonium nitrate, superphosphate and potassium chloride?

## The tillable land is 2,5 million ha. Calculate how much each fertilizer will we use in 1 ha. And how much fertilizers we use in total.

Mar 11, 2016

1 ha requires $\text{72.5 kg of N, 41.1 kg of P, and 40.4 kg of K}$ fertilizer.

The tillable land requires $\text{181 kt of N, 103 kt of P, and 101 kt of K}$ fertilizer.

#### Explanation:

Assumptions

• ${\text{NH"_4"NO}}_{3}$ fertilizer has an NPK rating 34-0-0 ($\text{34 % N}$).
• $\text{SSP}$ (single superphosphate) fertilizer has an NPK rating 0-20-0 (20 % "P"_2"O"_5)
• $\text{KCl}$ fertilizer has an NPK rating of 0-0-61 ($\text{61 % K"_2"O}$)

A small sample of the fertilizer with NPK in the ratio 3:1:3 might contain 3 kg of $\text{N}$,
1 kg of ${\text{P"_2"O}}_{5}$, and 3 kg of $\text{K"_2"O}$.

The mass of each fertilizer to that contains these amounts is:

$\text{Mass of 34-0-0 fertilizer" = 3 color(red)(cancel(color(black)("kg N"))) × "100 kg 34-0-0"/(34 color(red)(cancel(color(black)("kg N")))) = "8.82 kg 34-0-0}$

$\text{Mass of 0-20-0 fertilizer" = 1 color(red)(cancel(color(black)("kg P"_2"O"_5))) × "100 kg 0-20-0"/(20 color(red)(cancel(color(black)("kg P"_2"O"_5)))) = "5.00 kg 0-20-0}$

$\text{Mass of 0-0-61 fertilizer" = 3 color(red)(cancel(color(black)("kg K"_2"O"))) × "100 kg 0-0-61"/(61 color(red)(cancel(color(black)("kg K"_2"O")))) = "4.92 kg 0-0-61}$

$\text{Total mass of mixture"color(white)(l) =color(white)(l) "8.82 kg of 34-0-0}$
$\textcolor{w h i t e}{m m m m m m m m m m m m} \text{+ 5.00 kg of 0-20-0}$
$\textcolor{w h i t e}{m m m m m m m m m m m m} \text{+ 4.92 kg of 0-0-61}$
stackrel(color(white)(mmmmmmmmmmmmmmmmm)—————————)(color(white)(mmmmmmmmmll)"= 18.74 kg total")

So, we have 18.74 kg of fertilizer with a 3:1:3 ratio of $\text{N":"P"_2"O"_5:"K"_2"O}$.

In 154 kg of 3:1:3, there will be

154 color(red)(cancel(color(black)("kg 3:1:3"))) × "8.82 kg 34-0-0"/(18.74 color(red)(cancel(color(black)("kg 3:1:3")))) = "72.5 kg 34-0-0"

154 color(red)(cancel(color(black)("kg 3:1:3"))) × "5.00 kg 0-20-0"/(18.74 color(red)(cancel(color(black)("kg 3:1:3")))) = "41.1 kg 0-20-0"

154 color(red)(cancel(color(black)("kg 3:1:3"))) × "4.92 kg 0-0-61"/(18.74 color(red)(cancel(color(black)("kg 3:1:3")))) = "40.4 kg 0-0-61"

∴ 1 ha requires $\text{72.5 kg of 34-0-0, 41.1 kg of 0-20-0, and 40.4 kg of 0-0-61}$.

The tillable land requires

2.5 × 10^6 color(red)(cancel(color(black)("ha"))) ×" 72.5 kg 34-0-0"/(1 color(red)(cancel(color(black)("ha")))) = 1.81 × 10^8color(white)(l) "kg 34-0-0" = "181 kt 34-0-0"

2.5 × 10^6 color(red)(cancel(color(black)("ha"))) × "41.1 kg 0-20-0"/(1 color(red)(cancel(color(black)("ha")))) = 1.07 × 10^8color(white)(l) "kg 0-20-0" = "103 kt 0-20-0"

2.5 × 10^6 color(red)(cancel(color(black)("ha"))) × "40.4 kg 0-0-61"/(1 color(red)(cancel(color(black)("ha")))) = 1.81 × 10^8color(white)(l) "kg 0-0-61" = "101 kt 0-0-61"

It will require $\text{181 kt 34-0-0, 103 kt 0-20-0, and 101 kt 0-0-61}$ to fertilize the tillable land.