Calculate the amount of calcium oxide required when it reacts with `852 g` of `P_(4) O_(10)`.

Explanation

`underset("Base")(6CaO)+ underset("Acid")(P_(4)O_(10)) to underset("Salt")(3Ca_(3)(PO_(4))_(2))`
` "Moles of CaO"/("Moles of "P_(4)O_(10)) = 6/1`
[Mol wt. of `P_(4)O_(10) = 284`]
`rArr` Mole of CaO required ` = 6 xx " Mole of " P_(4)O_(10)`
`= 6 xx (852)/(284) `
Weight of CaO required ` = (6 xx 852)/284 xx 56 = 1008 ` g