"3 Ca(NO\u2083)\u2082 (aq) + 2 Li\u2083PO\u2084 (aq) \u2192" "Ca\u2083(PO\u2084)\u2082 (s) + 6 LiNO\u2083 (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"
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