How many moles of $\mathrm{CaO}$ are required to neutralize 142 g of ${\mathrm{P}}_4{\mathrm{O}}_{10}$

The reaction will be: 

$\begin{array}{l}6\mathrm{CaO}+{\mathrm{P}}_4{\mathrm{O}}_{10}\to 2{\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2\\ 142 \mathrm{g} {\mathrm{P}}_4{\mathrm{O}}_{10}\equiv 0.5 \mathrm{mol} {\mathrm{P}}_4{\mathrm{O}}_{10}\end{array}$

6 moles of CaO requires to neutralize one mole of P₄O₁₀ and 3 moles of CaO requires 0.5 moles of  P₄O₁₀ (or 142 g)