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mergl
1.34 grams of MgNH4PO4*6H2O
Molar mass= 245.1 grams per mole
Moles of MgNH4PO4*6H2O=
(1.34 grams)(1 mole/245.1 grams fertilizer)=0.0054671563 moles fertilizer
MgNH4PO4*6H2O has 1 mole of Phosphorous, so:
0.0054671563 moles fertilizer(1 mole Phosphorous/1 mole fertilizer)
0.0054671563 moles Phosphorous (30.97 grams/ 1 mole)
=0.1693178295 grams Phosphorous

[tex]\boxed{{\text{0}}{\text{.17}}\;{\text{g}}}[/tex]of phosphorus is present in 1.34 g of fertilizer.

Further Explanation:

Mole is a measure of the amount of substance. It is defined as the mass of a substance that has the same number of fundamental units as there are atoms in 12 g of carbon-12. Such fundamental units can be atoms, molecules or formula units.

The formula to calculate the moles of fertilizer [tex]\left({{\text{MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}}\right)[/tex]is as follows:

[tex]{\text{Moles of MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}=\frac{{{\text{Given mass of MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}}}{{{\text{Molar mass of MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}}}[/tex]                 …… (1)

The given mass of [tex]{\text{MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}[/tex] is 1.34 g.

The molar mass of [tex]{\text{MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}[/tex]is 245.1 g/mol.

Substitute these values in equation (1)

[tex]\begin{gathered}{\text{Moles of MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}=\left({1.34\;{\text{g}}}\right)\left({\frac{{1\;{\text{mol}}}}{{245.1\;{\text{g}}}}}\right)\\={\text{0}}{\text{.0054671563 mol}}\\\approx{\text{0}}{\text{.0055}}\;{\text{mol}}\\\end{gathered}[/tex]

The chemical formula of the fertilizer is [tex]{\text{MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}[/tex]. This indicates that one mole of the fertilizer contains one mole of phosphorus.

So the number of moles of phosphorus in 0.0054671563 moles of fertilizer can be calculated as follows:

[tex]\begin{gathered}{\text{Moles of phosphorus}}=\left({0.0054671563{\text{ mol MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}}\right)\left({\frac{{{\text{1 mol phosphorus}}}}{{{\text{1 mol MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}}}}\right)\\=0.0054671563{\text{ mol}}\\\end{gathered}[/tex]

The formula to calculate the mass of phosphorus is as follows:

[tex]{\text{Mass of phosphorus}}=\left({{\text{Moles of phosphorus}}}\right)\left({{\text{Molar mass of phosphorus}}}\right)[/tex] …… (2)

The number of moles of phosphorus is 0.0054671563 mol.

The molar mass of phosphorus is 30.97 g/mol.

Substitute these values in equation (2).

[tex]\begin{gathered}{\text{Mass of phosphorus}}=\left({{\text{0}}{\text{.0054671563 mol}}}\right)\left({\frac{{{\text{30}}{\text{.97 g}}}}{{{\text{1 mol}}}}}\right)\\={\text{0}}{\text{.1693178295 g}}\\\approx {\text{0}}{\text{.17 g}}\\\end{gathered}[/tex]  

So, 0.17 grams of phosphorus is present in the fertilizer.

Learn more:

1. Calculate the moles of chlorine: https://brainly.com/question/3064603

2. What is the number of moles of ions in HCl solution? https://brainly.com/question/5950133

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Mole concept

Keywords: moles, fertilizer, 0.17 g, MgNH4PO4.6H2O, molar mass, phosphorus, moles, mole, molar mass, 30.97 g/mol and0.0054671563 mol.

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