Question

2. 20.0 g of aqueous magnesium sulfate reacts with aqueous sodium phosphate and yields a soluble and an insoluble compound. (a) Write and balance the equation and indicate the physical state of reactants and products; (b) calculate the amount of insoluble product assuming more than enough sodium phosphate; and (c) find the number of moles of the insoluble product that can be obtained from 15.0 g sodium phosphate if there is more than enough of MgSO4. (25 points) (a) reaction (b) Grams of insoluble product from 20.0 g magnesium sulfate (c) Moles of insolouble product from 15.0 g sodium phosphate

Answer 1

2a) Reaction :

MgSO4 (aq) + Na3PO4 (aq) -> Mg3(PO4)2(s) + Na2SO4 (aq)

Balanced equation ;

3MgSO4(aq) + 2Na3PO4(aq) -> Mg3(PO4)2(s) + 3Na2SO4(aq)

b) 3 moles MgSO4 give 1 mole Mg3(PO4)2

3×120.66 g MgSO4 give 262.85 g Mg3(PO4)2

20 g MgSO4 give 262.85×20/3×120.66 g Mg3(PO4)2

= 14.52 g Mg3(PO4)2

c) 2 moles Na3PO4 give 1 mole Mg3(PO4)2

2×163.94 g Na3PO4 give 1 mole Mg3(PO4)2

15 g Na3PO4 give 15/2×163.94 moles Mg3(PO4)2

= 0.0457 moles