Question

How many grams of Ca₃(PO₄)₂ precipitate can form by reacting 258.6 mL of 1.8 M CaBr₂ with an excess amount of Li₃PO₄, given the balanced equation:

2 Li₃PO₄ (aq) + 3 CaBr₂ (aq) --> 6 LiBr₂ (aq) + Ca₃(PO₄)₂ (s)

Answer 1

Number of moles of CaBr2 = molarity * volume of solution in L

Number of moles of CaBr2 = 1.8 * 0.2586 = 0.465 mole

From the balanced equation we can say that

3 mole of CaBr2 produces 1 mole of Ca3(PO4)2 so

0.465 mole of CaBr2 will produce

= 0.465 mole of CaBr2 *(1 mole of Ca3(PO4)2 / 3 mole of CaBr2)

= 0.155 mole of Ca3(PO4)2

mass of 1 mole of Ca3(PO4)2 = 310.18 g so

the mass of 0.155 mole of Ca3(PO4)2 = 48.1 g

Therefore, the mass of Ca3(PO4)2 produced would be 48.1 g