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