Question

vula, Results and Discussion Beaker Sample 1 Name Mass of SrCl-6H,0 and container Mass of container Mass of SrCl, 6H 0 added to Beaker I 112.4444 110.2369 2. 2075 0.0048 Moles of SrC1,-6H 0 added to Beaker I Theoretical moles of Sr,(PO)2 determined from moles of SrC1,-6H,0 2. a. Mass of Na,PO, 12H,0 with container Mass of container b. 3.4H 116.2369 1. 2041 0.0032 d. Mass of Na, P0, 12H,0 added to Beaker I Moles of Na PO, 12H,0 added to beaker I Theoretical moles of Sr,(PO), determined from moles of Na PO, 12H,0 Theoretical moles of Sr,(PO), in Beaker 1 determined from the limiting reagent

Answer 1

3SrCl₂ + 2Na₃PO₄ = Sr₃(PO₄)₂ + 6 NaCl

Net chemcial reaction

So 3 moles of SrCl₂ produces 1 mole of Sr₃(PO₄)₂ and 2 moles of Na₃PO₄ produces 1 mole of Sr₃(PO₄)₂
We will calculate on this basis.

Moles of chloride .0048 So moles of strontium phosphate .0016

Moles of Sodium phosphate .0032 So moles of strontium phosphate .0016

Here Sodium phosphate is present in less amount .So it is the limiting reagent.

So base on limiting reagent theoretical yield is 0.0016 moles of Strontium phosphate.