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.

1 Answer
Mar 11, 2016

Answer:

1 ha requires #"72.5 kg of N, 41.1 kg of P, and 40.4 kg of K"# fertilizer.

The tillable land requires #"181 kt of N, 103 kt of P, and 101 kt of K"# fertilizer.

Explanation:

WARNING! Long Answer!

Assumptions

  • #"NH"_4"NO"_3# fertilizer has an NPK rating 34-0-0 (#"34 % N"#).
  • #"SSP"# (single superphosphate) fertilizer has an NPK rating 0-20-0 (#20 % "P"_2"O"_5#)
  • #"KCl"# fertilizer has an NPK rating of 0-0-61 (#"61 % K"_2"O"#)

A small sample of the fertilizer with NPK in the ratio 3:1:3 might contain 3 kg of #"N"#,
1 kg of #"P"_2"O"_5#, and 3 kg of #"K"_2"O"#.

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

#"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"#

#"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"#

#"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"#

#"Total mass of mixture"color(white)(l) =color(white)(l) "8.82 kg of 34-0-0"#
#color(white)(mmmmmmmmmmmm)"+ 5.00 kg of 0-20-0"#
#color(white)(mmmmmmmmmmmm)"+ 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 #"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 #"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 #"181 kt 34-0-0, 103 kt 0-20-0, and 101 kt 0-0-61"# to fertilize the tillable land.