How many grams of CaO are required to react with 852 g of P₄O₁₀?

$6\mathrm{CaO}+{\mathrm{P}}_4{\mathrm{O}}_{10}\to 2{\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$
1 mole of P₄O₁₀ = molar mass of P₄O₁₀ = 284 g
852 g of P₄O₁₀ $=\frac{852}{284}=3 \mathrm{moles}$
1 mole of P₄O₁₀ reacts with 6 moles of CaO
3 moles of P₄O₁₀ reacts with 18 moles of CaO
Mass of 18 moles of CaO = 18 x 56 = 1008 g