contestada

A multi-nutrient fertilizer contains several different nitrogen containing compounds. The fertilizer is 54.8% CH4N2O (urea), 26.3% KNO3 , and 14.1% (NH4)2HPO4 by mass. The remainder of the fertilizer consists of substances that do not contain nitrogen. How much fertilizer should someone apply to provide 2.90 g N to a plant?

Respuesta :

Edufirst

Answer:

  • 9.01 g of fertilizer

Explanation:

Assume initially that the amount in grams of the multinutrient fertilizer is 100 g, and calculate the amount of nitrogen supplied by every compound present in these 100 g of fertilizer.

1) Urea (CH₄N₂O):

  • Molar mass of urea: 60.06 g/mol
  • Atomic mass of N: 14.007 g/mol
  • Total mass of N in the formula: 2 × 14.007 g/mol = 28.014 g/mol
  • Amount of N in 100 g of compound: 100 g × 54.8% × 28.014 g / 60.06 g =  25.56 g

2) KNO₃

  • Molar mass of KNO₃: 101.1032 g/mol
  • Atomic mass of N: 14.007 g/ mol
  • Total mass of N in the formula: 14.007 g/mol
  • Amount of N in 100 of the compound: 100 g × 26.3% × 14.007 / 101.1032 = 3.64 g

3) (NH₄)₂PO₄

  • Molar mass of (NH₄)₂PO₄: 132.06 g/mol
  • Atomic mass of N: 14.007 g/ mol
  • Total mass of N in the formula: 2×14.007 g/mol = 28.014 g/mol
  • Amount of N in 100 of the compound: 100 g × 14.1% × 14.007 / 132.06 = 2.99 g

4) Total mass of N in 100 g of fertilizer:

Add all the amounts of N obtained above

  • 25.56g + 3.69 g + 2.99 g = 32.19 g of N

5) Mass of fertilizer that should be applied to provide 2.90 g of N to a plant.

Set a proportion:

  • 32.19 g of N / 100 g of fertilizer = 2.90 g of N / X

Solve for X:

  • X = 2.90 g of N × 100 g of fertilizer / 32.19 g of N = 9.01 g of fertilizer

That is the answer: 9.01 g of fertilizer should be applied to provide 2.90g  of N to a plant.

lhmarianateixeira

In this exercise we have to use proportionality and knowledge in chemistry to calculate how much fertilizer should be applied to the plant, so we have to:

9.01 g of fertilizer

First we have to score the known values ​​before starting the calculations so we have:

  • Molar mass of urea: 60.06 g/mol
  • Atomic mass of N: 14.007 g/mol
  • Total mass of N in the formula: 28.014 g/mol

  • Molar mass of KNO₃: 101.1032 g/mol
  • Atomic mass of N: 14.007 g/ mol
  • Total mass of N in the formula: 14.007 g/mol

  • Molar mass of (NH₄)₂PO₄: 132.06 g/mol
  • Atomic mass of N: 14.007 g/ mol
  • Total mass of N in the formula: 28.014 g/mol

Are performing the proportionality of urea we find that:

[tex]100 g * 54.8\% * 28.014 g / 60.06 g = 25.56 g[/tex]

Are performing the proportionality of  KNO₃ we find that:

[tex]100 g * 26.3\% * 14.007 / 101.1032 = 3.64 g[/tex]

Are performing the proportionality of (NH₄)₂PO₄ we find that:

[tex]100 g * 14.1\% * 14.007 / 132.06 = 2.99 g[/tex]

Total mass of N in 100 g of fertilizer:

[tex]25.56g + 3.69 g + 2.99 g = 32.19 g[/tex]

Mass of fertilizer that should be applied is:

[tex]32.19 g / 100 g = 2.90 g of N / X\\X = 2.90 * 100 g / 32.19g = 9.01 g[/tex]

See more about proportionality at brainly.com/question/2548537