Question

A student uses 3.00g of a fertilizer with an npk ratio of 15-15-10, how much mgnh4po4*6h2o (mm =245.1 g/mol) should the student produced?

Answer 1

The correct answer is 8.89 g.

The production of MgNH₄PO₄.6H₂O is given by the equation:

NH₃ + H₃PO₄ ⇒ (NH₄)H₂PO₄

(NH₄)H₂PO₄ + MgSO₄ + 6H₂O ⇒ MgNH₄PO₄.6H₂O + H₂SO₄

This is the real fertilizer of NPK of weight 3 g.

It is given that the NPK is present in the ratio of 15-15-10.

Therefore, the moles of phosphorus found in the actual fertilizer is as follows:

% of P present as P₂O₅ in the fertilizer is (15/40) × 100 = 37.5 %

% of P = mass of P / actual mass of fertilizer

0.375 × 3 = mass of P

Mass of P = 1.125 g / 31

Moles of P = 0.03629 mol

On the basis of reaction stoichiometry,

1 mol of P in the actual fertilizer = 1 mol of P in the product

0.03629 of P

Moles of MgNH₄PO₄.6H₂O = 0.03629 mol

Mass of MgNH₄PO₄.6H₂O formed = 0.03629 mol × 245.1 g/mol

= 8.89 g

Thus, the mass of the product MgNH₄PO₄.6H₂O formed is 8.89 g.

Answer 2

The amount of ${\text{ MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot 6{{\text{H}}_2}{\text{O}}$  produced is $\boxed{{\text{3}}{\text{.5657 g}}}$ .

Further explanation:

Stoichiometry of a reaction is used to determine the amount of species present in the reaction by the relationship between the reactants and products. It can be used to determine the moles of a chemical species when the moles of other chemical species present in the reaction is given.

Consider the general reaction,

${\text{A}}+2{\text{B}}\to3{\text{C}}$

Here,

A and B are reactants.

C is the product.

One mole of A reacts with two moles of B to produce three moles of C. Stoichiometric ratio between A and B is 1:2, the stoichiometric ratio between A and C is 1:3 and the stoichiometric ratio between B and C is 2:3.

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 NPK ratio of 15-15-10. This implies the percentage of nitrogen in the fertilizer is 15 %, that of phosphorus is 15 % and that of potassium is 10 %.

P is the symbol for phosphorus.

The formula to calculate % of P in fertilizer is written as follows:

${\text{\% of P}}=\frac{{{\text{mass of P}}}}{{{\text{mass of fertilizer}}}}$                                  …… (1)

Rearrange equation (1) to calculate mass of P.

${\text{mass of P}}={\text{\% of P}}\times{\text{mass of fertilizer}}$   …… (2).

Given the mass of fertilizer is 3.00 g.

% of P is 15%.

Substitute these values in equation (2).

$\begin{aligned}{\text{mass of P}}&=\frac{{15}}{{100}}\left({{\text{3}}{\text{.00 g}}}\right)\\&={\text{0}}{\text{.45 g}}\\\end{aligned}$

The formula to calculate number of moles of P in fertilizer is as follows:

${\text{Number of moles }}=\frac{{{\text{mass}}}}{{{\text{molar mass}}}}$                                  ……(3)

Molar mass of P is 30.97g/mol.

Mass of P is 0.45 g.

Substitute these values in equation (3).

$\begin{aligned}{\text{Number of moles}}&=\frac{{{\text{0}}{\text{.45 g}}}}{{{\text{30}}{\text{.97 g/mol}}}}\\&={\text{0}}{\text{.01453 mol}}\\\end{aligned}$

Hence, the number of moles of P in fertilizer is 0.014530 mol.

The reaction for production of ${\text{MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}$ is as follows:

$\begin{gathered}{\text{N}}{{\text{H}}_3}+{{\text{H}}_3}{\text{P}}{{\text{O}}_4}\to\left({{\text{N}}{{\text{H}}_4}}\right){{\text{H}}_2}{\text{P}}{{\text{O}}_4}\hfill\\\left({{\text{N}}{{\text{H}}_4}}\right){{\text{H}}_2}{\text{P}}{{\text{O}}_4}+{\text{MgS}}{{\text{O}}_4}+{\text{6}}{{\text{H}}_2}{\text{O}}\to{\text{MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}+{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\hfill\\\end{gathered}$

According to stoichiometry of the reaction, 1 mole of P in the fertilizer reacts to form 1 mole of P in product.  

Therefore number of moles of ${\text{MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}$  will be equal to number of moles of P in fertilizer and that is0.01453 mol.

Molar mass of ${\text{MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}$ is 245.1 g/mol.

Rearrange equation (3) to calculate mass of ${\text{MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}$ .

$\begin{aligned}{\text{mass}}&={\text{number of moles}}\times{\text{molar mass}}\\&={\text{0}}{\text{.01453 mol}}\left({\frac{{{\text{245}}{\text{.40 g}}}}{{{\text{mol}}}}} \right)\\&={\text{3}}{\text{.565708 g}}\\&\approx{\text{3}}{\text{.5657 g}}\\\end{aligned}$

The amount of ${\text{MgN}}{{\text{H}}_4}{\text{P}}{{\text{O}}_4}\cdot6{{\text{H}}_2}{\text{O}}$ produced is 3.5657g.

Learn more:

1. The neutral element represented by the excited state electronic configuration: brainly.com/question/9616334

2. Calculation of equilibrium constant of pure water at 25°c: brainly.com/question/3467841

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Environmental chemistry

Keywords: Stoichiometry, reactant, product, number of moles, fertilizers, N-P-K ratio,15%,10%, molar mass,3.5613 g.