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.
