How much Ca3(PO4)2(s) could be produced in an industrial process if 55.00 g of CaCl2 in solution reacted completely with sufficient Na3(PO4) (aq)

51.1 g of  Ca₃(PO₄)₂(s) can be made in this reaction

Explanation:

The reactans are CaCl₂ and Na₃PO₄. Let's determine the reaction:

3CaCl₂(aq)  +  2Na₃PO₄(aq)  →  Ca₃(PO₄)₂(s) ↓ + 6NaCl(aq)

We convert the mass of chloride to moles:

55 g . 1 mol/ 110.98 g = 0.495 moles

Ratio is 3:1. Let's make a rule of three to find the answer in moles:

3 moles of chloride can produce 1 mol of phosphate

Therefore 0.495 moles will produce (0.495 . 1) / 3 = 0.165 moles

We convert the moles to mass:

0.165 mol .  310.18 g /1 mol = 51.1 g

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