Astudent 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?

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.

the answer would be b i believe