Question

How many moles of precipitate are formed when 24.0 mL of 0.250 M Ca(NO₃)₂ is mixed with excess Li₃PO₄ in the following chemical reaction?
3 Ca(NO₃)₂ (aq) + 2 Li₃PO₄ (aq) → Ca₃(PO₄)₂ (s) + 6 LiNO₃ (aq)

Answer 1

$3 Ca(NO₃)₂ (aq) + 2 Li₃PO₄ (aq) →$ $Ca₃(PO₄)₂ (s) + 6 LiNO₃ (aq)$

As $Li_3PO_4$ is in excess,reaction is independent of its concentration.

As per the reaction,$3$ moles of calcium nitrate produce $1$ mole of calcium phosphate or

The precipitate formed will be one-third of the reactant.

Moles of $Ca(NO_3)_2=M\times V (in\ L)=$

$0.25\times24\times 10^{-3}=0.006\ mol$

Moles of $Ca_3(PO_4)_2$ formed $=\frac{1}{3}\times 0.006=0.002\ mol$