Question

How many grams of Ca3(PO4)2 precipitate can form by reacting 325.2 mL of 1.6 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

Balanced equation for the given reaction is,

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

We know that moalrity of a solution is given by,

Molarity = No. of moles of substance / Volume of solution

Also, we are given

Volume of CaBr₂ solution = 325.2 ml = 0.3252 L              

Molarity of CaBr₂ solution = 1.6 mol / L                                                     

On substituting given values we get,

No. of moles of CaBr₂ in solution = Volume * Molarity of Solution

No. of moles of CaBr₂ in solution = (0.3252 L) * (1.60 mol/L) = 0.5203 mol

According to the coeffecient ratio rule we know that in given reaction, CaBr₂ and Ca₃(PO₄)₂ react in a ratio of 3 : 1 i.e. for every 3 moles of CaBr₂ 1 mole of Ca₃(PO₄)₂ is required.

No. of moles of CaBr₂ in solution = 0.5203 mol

Thus, no of moles of Ca₃(PO₄)₂ precipitate in reaction = (0.5203) / 3 = 0.17344 mol

Molar mass of Ca₃(PO₄)₂ = 310 g/mol

Thus, mass of Ca₃(PO₄)₂ precipitate in reaction = ( 0.17344 mol ) * ( 310 g/mol )

No. of moles of Ca₃(PO₄)₂ precipitate in reaction = 53.7664 g = 53.77 g