Question

1.Calculate the number of moles of phosphorus (P) in 5.64 g of a 15-30-15 fertilizer.

2.Determine the volume in milliliters of 0.400 M MgSO₄ solution that must be added to completely react with 0.02549 moles of phosphorus present in a fertilizer sample.

Answer 1

1) 15-30-15 fertilizer means N-P-K in order

Mass of P = (30/15+30+15) x10= 5.0 g

molar mass of P = 30.97

Moles of P= mass of P / molar mass of P

                = 5.0/30.97

                = 0.1614

moles of P = 0.1614

2)

H3PO4 moles = P moles = 0.02549

MgSO4 moles = 0.400 x V

3 MgSO4 + 2 H3PO4 ----------------------> Mg3(PO4)2 + 3 H2SO4

3 mol MgSO4 ---------------------> 2 mol H3PO4

0.400V mol MgSO4 -------------> 0.02549 mol H3PO4

2 x 0.400 V = 3 x 0.02549

V = 0.09559 L

V = 95.59 mL

volume of MgSO4 needed = 95.59 mL