A) 5Ca(OH)₂ + 3H₃PO₄ → Ca₅(PO₄)₃(OH) + 9H₂O
B) m=158 g; w=0.830; m{Ca(OH)₂}=123 g
n(H₃PO₄) = m(H₃PO₄)/M(H₃PO₄) = mw/M(H₃PO₄)
n(H₃PO₄) = 158g*0.830/(98.0g/mol) = 1.3382 mol
n{Ca(OH)₂}=m{Ca(OH)₂}/M{Ca(OH)₂}
n{Ca(OH)₂}=123g/(74.01g/mol)=1.6619 mol
Ca(OH)₂:H₃PO₄ = 5:3
1.6619:1.3382 = 5:4 the limiting reagent is a calcium hydroxide
m{Ca₅(PO₄)₃(OH)} = M{Ca₅(PO₄)₃(OH)}n{Ca(OH)₂}/5
m{Ca₅(PO₄)₃(OH)} = 502.3g/mol*1.6619/5=166.954 g*
*The statement of the problem is incorrect, surplus of a phosphoric acid reacts with a hydroxyapatite. The calculated quantity of a hydroxyapatite is not formed.
3Ca₅(PO₄)₃OH + H₃PO₄ = 5Ca₃(PO₄)₂ + 3H₂O
